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[DOC File]AP CHEMISTRY
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Given the following equation: H2 + I2 ( 2 HI (all gases) and the Keq = 710 . If there is 15.7 grams of Hydrogen gas and 294 grams of iodine in a 5 liter flask, what will be the equilibrium concentration for all three materials ? ... The acid HOCl has a pKa value of 7.50 . Calculate the pH of a solution containing 0.25 M HOCl and 0.75 M NaOCl.
direction of equilibrium

[DOC File]Chapter 9
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In the equation above Keq is the “equilibrium constant,” (1, pg. 256). It can be used to tell “whether the products or the reactants are favored,” (1, pg. 256) at equilibrium. Which group is favored is easily figured out by the below equation. ... The equation illustrating this is below. pKa + pKb = 14.00.
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[DOC File]Lab 3 - pH and Buffer Lab - Arkansas State University
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Use “p” notation: pH = pKa + log ([CH3COO-]/CH3COOH]) This is the Henderson-Hasselbach equation, which relates the pKa of a solution (a measure of how easily an acid gives up its H+) to the pH of a solution of that acid. Note, however, that the initial concentrations of the weak acid and its conjugate base are also important to the equation.
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[DOC File]Understanding the shapes of titration curves
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Write the balanced equation for each step along with their equilibrium constants. Add the 3 reactions and calculate the equilibrium constant for the net reaction. Step 1: Keq = Step 2: Keq = Step 3: Keq = Net: Keq = ... label the region in which the solution is a buffer and the point where the pH = pKa. Titration of 25.00 mL of 0.100 M HCl with ...
keq 1

[DOC File]Acid - Base Balance: Chapter 18; pages 583 - 595
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Henderson-Hasselbalch Equation [H+] = Keq x [HA] [A-]-log [H+] = -log Keq + log [A-] [HA] pH = pKa + log [A-] [HA] pKa. when pH = pK, acid is 50% dissociated. when pH = pK, 50% of buffer is in each form giving maximum buffering capacity. match pK of potential buffers to desired pH. Metabolism.
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[DOC File]Chemical equilibrium worksheet A (answer key)
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(This problem initially appears to be an unsolvable two variable equation, until you actually evaluate the Keq) Keq= 0.00237 M-1 = [NO2]2 / [NO] [O2]2 = X2 / YX2 (The x’s cancel out) Y= [O2] = 422 M. equilibrium lies right, products favored. 5) decomposition of diatomic iodine to monoatomic iodine.
pka equilibrium

Keq equation pka