ࡱ> '`0RbjbjLULU :.?.?J& 4 |-|-|-8-p.L4 -j.Z4"|4|4|4W5W5W5$h҅ 6W5W566҅ |4|4;;;6 |4 |4;6;;y (||4. @{r'|-8{60-{6:l(|(| W5Z5@;54%6W5W5W5҅҅k;XW5W5W5-66664 4 4 Dx4 4 4 x4 4 4    HYPERLINK "http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-strong-acid-q1" What is pH of 0.0002M HI solution? What will be pH if 1mL of the above was added to 999mL of pure water?  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-of-mixture-q1" What is pH of mixture prepared by mixing 20mL 0.07M NaOH and 13mL 0.09M HCl?  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-weak-acid-q1" Calculate pH of 1M solution of nitrous acid, pKa=3.37.  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-weak-acid-q2" Calculate pH of 0.001M solution of benzoic acid, pKa=4.19.  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation-questions&right=weak-acid-pKa-q1" 0.1M solution of weak acid has pH=4.0. Calculate pKa.  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation-questions&right=weak-acid-pKa-q2" 20mL of 0.1M solution of weak acid was mixed with 8mL 0.1M solution of NaOH. Measured pH was 5.12. Calculate pKa.  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-weak-base-q1" Calculate the pH of a 0.2M solution of ammonia. Ka=5.6210-10.  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-weak-base-hydrolysis-q1" Calculate pH of 0.01M aniline hydrochloride. Aniline pKb=9.4.  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-amphiprotic-salt-q1" Calculate pH of 0.1M potassium hydrogen oxalate. pKa1=1.25, pKa2=4.27.  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-buffer-q1" Calculate pH of solution prepared from 0.1mole of formic acid and 0.02mole of NaOH diluted to 1L. PKa=3.75.  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-buffer-q2" You have 100mL of 1M ammonia solution (pKa=9.25). What volume of 1M hydrochloric acid is needed to prepare  What is pH of 0.0002M HI solution? What will be pH if 1 mL of the above was added to 999 mL of pure water? HI is a strong acid so we can assume it is fully dissociated. 310-4M solution is concentrated enough to allow us to neglect water dissociation (see table in  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-strong-acid-base" \l "hcltable" strong acid section). Thus we can use  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-strong-acid-base" \l "eq7.6" 7.6 equation  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/7-6.png" \* MERGEFORMATINET 7.6 and the pH calculated is -log(310-4)=3.70. After the solution has been diluted 1000 times HI concentration is 210-7M and we are in the area where water dissociation may spoil results of the  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-strong-acid-base" \l "eq7.6" 7.6 equation. Thus it wil be better to use the equation  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-strong-acid-base" \l "eq7.4" 7.4  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/7-4.png" \* MERGEFORMATINET 7.4 and calculated pH=6.62 (you know how to solve quadratic equation, don't you?). Using equation  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-strong-acid-base" \l "eq7.6" 7.6 we will get pH=6.70. Difference is not large - but clearly visible. As additional confirmation our  HYPERLINK "http://www.chembuddy.com/?left=BATE&right=pH-calculator" pH calculator shows exactly the same results - 3.70 and 6.62. What is pH of mixture prepared by mixing 20mL 0.07M NaOH and 13mL 0.09M HCl? Although you are asked about pH, this is a simple limiting reagent question. When you mix strong acid and strong base they react till one of the reagents runs out. What is left is responsible for pH. Reaction taking place in the solution is NaOH + HCl -> NaCl + H2O Initially there were 200.07=1.4mmol of NaOH and 130.09=1.17mmol of HCl. They react 1:1, so after reaction 1.4-1.17=0.23mmol of NaOH is left. Final volume of the solution is 20+13=33mL, thus final concentration of NaOH left is 0.23/33=0.0070M. pOH=-log(0.0070)=2.15 and pH=14-2.15=11.85. Note that if there is exactly the same amount of both reactants reaction proceeds to the end - and you are left with perfectly neutral solution of pH=7.00. Calculate pH of 1M solution of nitrous acid, pKa=3.37. Let's try first with the simplified formula  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base" \l "eq8.13" 8.13.  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/8-13.png" \* MERGEFORMATINET 8.13 If we put concentration and Ka values into formula we get [H+]=0.021M and pH=1.69. Is it correct? To be sure we have to check if the  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=general-pH-calculation" \l "5-percent-rule" 5% rule is obeyed, that is if the dissociation fraction is less then 5%.  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/1-3.png" \* MERGEFORMATINET 1.3 Dissociated acid concentration is the same as [H+] - if we assume we can neglect water dissociation. It is obvious that we can, as 0.021M is five orders of magnitude larger than the concentration of H+ from the water. So dissociation fraction is 0.021/1 - a little bit over 2%, much less then 5%, so the 5% rule is obeyed. As additional confirmation our  HYPERLINK "http://www.chembuddy.com/?left=BATE&right=pH-calculator" pH calculat Calculate pH of 0.001M solution of benzoic acid, pKa=4.19. We will start with the simplified formula  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base" \l "eq8.13" 8.13.  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/8-13.png" \* MERGEFORMATINET 8.13 From formula we get [H+]=2.5410-4M and pH=3.60. Is it correct? To be sure we have to check if the  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=general-pH-calculation" \l "5-percent-rule" 5% rule is obeyed, that is if the dissociation fraction is less then 5%.  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/1-3.png" \* MERGEFORMATINET 1.3 Dissociated acid concentration is the same as [H+], thus dissociation fraction is 2.5410-4/0.001 - above 25%. We can't use simplified formula  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base" \l "eq8.13" 8.13! We have to use more precise equation  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base" \l "eq8.10" 8.10:  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/8-10.png" \* MERGEFORMATINET 8.10 This time we get pH=3.65, exactly the 0.1M solution of weak acid has pH=4.0. Calculate pKa. If pH=4.0 then [H+]=10-4 - that means that dissociation fraction is about 10-4/0.1100%=0.1% - well below 5%. That in turn means that pH of the solution is described by the simplified formula  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base" \l "eq8.13" 8.13.  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/8-13.png" \* MERGEFORMATINET 8.13 All we have to do is rearrange to get equation for Ka:  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/8-12.png" \* MERGEFORMATINET  and Ka=(10-4)2/0.1=10-7 and pKa=7. Our  HYPERLINK "http://www.chembuddy.com/?left=BATE&right=pH-calculator" pH calculator can help us check this result - and indeed, entering 0.1 as concentration and 7 as pKa we get pH=4.0 which confirms our calculations. 20mL of 0.1M solution of weak acid was mixed with 8 mL 0.1M solution of NaOH. Measured pH was 5.12. Calculate pKa. In the case of not-so-weak acids you can assume the neutralization reaction is quantitative (all of the strong base reacts with the weak acid) and the pH of the solution is described by  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch" \l "eq15.2" Henderson-Hasselbalch equation:  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/15-2.png" \* MERGEFORMATINET 15.2 All you have to do is to find out concentrations of A- and HA. However, it is worth of noting here, that you can save part of the calculations, replacing concentrations with numbers of moles - volume is the same for both substances so it cancels out. Solution was prepared using 200.1=2 mmol of acid and 80.1=0.8 mmol of strong base. If the reaction was quantitative after the neutralization there was 1.2 mmol of acid HA left and 0.8 mmol of A-. Inserting these values into Henderson-Hasselbalch equation gives 5.12=pKa+log(0.8/1.2). Solving for pKa you get pKa=5.12-log(0.8/1.2)=5.12+0.18=5.30.  HYPERLINK "http://www.chembuddy.com/?left=BATE&right=pH-calculator" pH calculator can help you check this result. Enter 5.3 as pKa, 0.1 as concentration of acid, select sodium hydroxide as a base, enter 0.1 as its concentration, finally select Ca+Cb+Va+Vb as a way of expressing solution composition and enter 20 and 8 as acid and base solutions volumes - and the pH displayed will be 5.12, confirming the result. Note that for strong or very weak acids and/or for very diluted solutions this method may give wrong results - when pH is further changed by dissociation or hydrolyzis or when the water autodissociation can't be neglected. Calculate the pH of a 0.2M solution of ammonia. Ka=5.6210-10. Ammonia is a weak base, so the most convenient approach is to calculate pOH using Kb, and then to convert it to pH. From the  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=bronsted-lowry-theory" \l "eq2.4" Brnsted-Lowry theory we know that KaKb=Kw, that allows us to calculate Kb=Kw/Ka=10-14/5.6210-10=1.7810-5. Calculation of the pOH will be now identical with the calculation of pH of a weak acid. Let's try first with the simplified formula  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base" \l "eq8.17" 8.17:  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/8-17.png" \* MERGEFORMATINET 8.17 If we put concentration and Kb values into formula we get [OH+]=0.00189M and pOH=2.72. Is it correct? To be sure we have to check if the  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=general-pH-calculation" \l "5-percent-rule" 5% rule is obeyed, that is if the dissociation fraction is less then 5%.  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/1-3.png" \* MERGEFORMATINET 1.3 Dissociated base concentration is the same as [OH-] - if we assume that we can neglect water dissociation. It is obvious that we can, as 0.00189M is about five orders of magnitude larger than the concentration of OH- from the water. So dissociation fraction is 0.00189/0.2 - below 1%, much less then 5%, so the 5% rule is obeyed. Now we need the final touch - we know pOH=2.72, but we need pH. Time to use equation  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product" \l "eq1.6" 1.6. pH=14-2.72=11.28. Our  HYPERLINK "http://www.chembuddy.com/?left=BATE&right=pH-calculator" pH calculator shows 11.27. Difference in the last digit can be probably attributed to different dissociation constant used. Calculate pH of 0.01M aniline hydrochloride. Aniline pKb=9.4. Aniline in solution of its hydrochloride is in form of conjugate acid. To calculate pH of such solution we have to find pKa for aniline and treat it like weak acid. Knowing pKb we can find pKa from the equation  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=bronsted-lowry-theory" \l "eq2.5" 2.5:  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/2-5.png" \* MERGEFORMATINET 2.5 pKa=14-9.4=4.6 - and this value we will put into simplified formula  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base" \l "eq8.13" 8.13.  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/8-13.png" \* MERGEFORMATINET 8.13 Result is [H+]=5.010-4M and pH=3.30. We have used simplified formula, so now we have to check if the  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=general-pH-calculation" \l "5-percent-rule" 5% rule is obeyed:  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/1-3.png" \* MERGEFORMATINET 1.3 Dissociation fraction is 5.010-4/0.01 - exactly 5%. We were allowed to use equation  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base" \l "eq8.13" 8.13. Our  HYPERLINK "http://www.chembuddy.com/?left=BATE&right=pH-calculator" pH calculator doesn't calculate pH for salts, but salt is nothing else but equimolar solution of acid and base. BATE calculates pH as 3.31 - where is the source of difference? First suspect is 5% rule - it doesn't state results will be exact, but the error will be acceptable. Let's try to calculate the result using more precise formula  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base" \l "eq8.10" 8.10.  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/8-10.png" \* MERGEFORMATINET 8.10 This time [H+]=4.8910-4 and pH=3.31. This result is more precise - but is it better and worth additional calculations? No! Dissociation constant is given only with two significant digits accuracy, so our pH is in fact 3.3 - or even just 3, as the concentration is given with only one significant digit.  HYPERLINK "http://www.chembuddy.com/?left=BATE&right=pH-calculator" pH calculator always calculate the result with full precision - that's what computers are for. It is up to us - humans - to understand the results. Calculate pH of 0.1M potassium hydrogen oxalate. pKa1=1.25, pKa2=4.27. This question - if done by hand - requires either approximate method (that gives questionable results) or tedious calculations. In  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt" amphiprotic salt section we have derived equation  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt" \l "eq12.9" 12.9  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/12-9.png" \* MERGEFORMATINET 12.9 According to the analysis we have done (results shown in the  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt" \l "table2" table) equation  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt" \l "eq12.9" 12.9 gives good results for more concentrated solutions. 0.01M is a little bit too diluted for the precise result. PH=1/2(1.25+4.27 ) =2.76. Let's check this result using  HYPERLINK "http://www.chembuddy.com/?left=BATE&right=pH-calculator" pH calculator. To create potassium hydrogen oxalate solution you have to select oxalic acid and potassium hydroxide and to enter 0.1 as both concentrations.  HYPERLINK "http://www.chembuddy.com/?left=BATE&right=pH-calculator" pH calculator shows 2.86 - and this is the correct result. Approximate method is very fast - but gives only very rough results. Calculate pH of solution prepared from 0.1mole of formic acid and 0.02mole of NaOH diluted to 1L. pKa=3.75. This is classic buffer question that we will solve using  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch" \l "eq15.2" Henderson-Hasselbalch equation.  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/15-2.png" \* MERGEFORMATINET 15.2 We need to know concentration of HCOO- and HCOOH. Formic acid has pKa value larger then 2.5 (see  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch" pH of buffer section) so we can assume stoichiometry is solely responsible for concentrations of both acid and conjugate base. If so, after neutralization has taken place [HCOOH]=0.1-0.02=0.08M and [HCOO-]=0.02M. Putting these values into Henderson-Hasselbalch equation we get: pH=3.75+log(0.02/0.08)=3.15. Using  HYPERLINK "http://www.chembuddy.com/?left=BATE&right=pH-calculator" pH calculator we get 3.17. Our rule of thumb (the one about pKa>=2.5) states that difference between real pH and calculated pH will be acceptable, not zero. You have 100mL of 1M ammonia solution (pKa=9.25). What volume of 1M hydrochloric acid is needed to prepare buffer with pH=9.5? Adding hydrochloric acid to the solution of ammonia (base) we create a conjugate acid NH4+. Ratio of their concentrations at pH 9.5 will be given by the  HYPERLINK "http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch" \l "eq15.2" Henderson-Hasselbalch equation.  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/15-2.png" \* MERGEFORMATINET 15.2 If all added acid reacts with ammonia, amounts of conjugate acid NH4+ created and base NH3 left is defined by the stoichiometry of the neutralization reaction:  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/qbq2-1.png" \* MERGEFORMATINET qbq2.1 and  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/qbq2-2.png" \* MERGEFORMATINET qbq2.2  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/qbq2-3.png" \* MERGEFORMATINET qbq2.3 qbq2.3 is equivalent to  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/qbq2-4.png" \* MERGEFORMATINET qbq2.4 where na and nb are number of moles of acid and conjugate base, nNH3start is number of moles of ammonia at the beginning (0.1 mole - n=CV) and nHCl is number of moles of acid added. We are using number of moles as it makes calculation much easier - this way we don't have to calculate how the concentrations change due to the dilution once the acid is added to original ammonia solution. We can do that thanks to the fact that in the Henderson-Hasselbalch equation volumes can be canceled out and replaced by mole ratio. Putting pH=9.5 into the Henderson-Hasselbalch equation we get  INCLUDEPICTURE "http://www.chembuddy.com/pH-calc-img/qbq2-5.png" \* MERGEFORMATINET qbq2.5 Finally we have a set of equations qbq2.4 and qbq2.5. When solved they give na=0.036 and nb=0.064. From qbq2.2 we get that we have to add 0.036 mole of HCl, or 36mL of 1M solution. We can use  HYPERLINK "http://www.chembuddy.com/?left=BATE&right=pH-calculator" pH calculator to check the result. Create a solution containing both ammonia and hydrochloric acid. Enter 1 as both concentrations. From the drop down list above pH sign select Ca+Cb+Va+Vb. Enter 36 as acid volume and 100 as base volume. Perfect! Note: if you need program that will help in buffer calculation, our  HYPERLINK "http://www.chembuddy.com/?left=BATE&right=pH-calculator" pH calculator is not suited for the task, however, you can try our  HYPERLINK "http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator" Buffer Maker - the  HYPERLINK "http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator" buffer calculator.  HYPERLINK "http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator" Buffer calculator was specifically designed to help in buffer preparation, so - given pH - it will automatically calculate recipe of any buffer you may need.     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