Calculate the pH of a 0.25 M solution of HCN. Calculate the pH of a solution tha

Question

Calculate the pH of a 0.25 M solution of HCN. Calculate the pH of a solution that is 0.35 M NaCN and 0.25 M HCN. Calculate the pH after adding 0.050 mol of HC1 to 1.00 L the solution in part b. A 15.0 mL sample of 0.200 M nitrous acid solution is titrated with a 0.100 M KOH solution. Calculate the volume of KOH solution required to reach the equivalence point. Calculate the pH of the sample at the equivalence point. Choose an indicator for the titration and briefly justify your choice. Calculate the pH after the addition of 12.0 mL of the KOH solution.

Explanation / Answer

5a.

pH of a Monoprotic Acid

By definition, in equilibrium:

Ka = [H+][A-]/[HA]

Then

Assume that [H+] = [A-] = x … due to stoichiometry (i.e. 1 mol of H+ per each mol of A-)

Then, in equilibrium [HA] = M – x (i.e. the original concentration minus the acid in equilibrium)

Substitute

Ka = (x*x)/(M-x)

3.2*10^-10 = x*x/(0.25-x)

solve for x (quadratic equation)

x = [H+] = 8.94*10^-6 M

[H+] = [A-] = x = 8.94*10^-6

pH = -log([H+]) =-log( 8.94*10^-6) = 5.048

b.

this is a buffer so

pH = pKa + log(A-/HA)

pH =9.21 + log(0.35/0.25) = 9.356

after adding 0.05 M

[conjguate] = 0.35 -0.05 = 0.30

[Acid = 0.20 +0.05= 0.25

pH =9.21 + log(0.35/0.25) = 9.3561

NOTE Consider posting multiple questions in mutliple set of Q&A.

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