A 100.0 mL sample of a solution that is 0.106 M in HCl and 0.106 M in HCN is tit

Question

A 100.0 mL sample of a solution that is 0.106 M in HCl and 0.106 M in HCN is titrated with 0.106 M NaOH. Calculate the pH after the addition of the following volumes of NaOH. Part A 0.0 mL Express your answer using three decimal places. pH = SubmitMy AnswersGive Up Part B 75.0 mL Express your answer using three decimal places. pH = SubmitMy AnswersGive Up Part C 100.0 mL Express your answer using three decimal places. pH = SubmitMy AnswersGive Up Part D 127 mL Express your answer using three decimal places.

Explanation / Answer

HCl is a strong acid and it will be completely ionized in the solution. But HCN is a weak acid and will not be completely ionize.

Concentration of [H+] from HCl = 0.106 M = 0.106 M *0.1L = 0.0106 moles

Dissociation constant of HCN = 6.2 x 10^-10

Ka = [H+][CN-]/[HCN]

or, 6.2 x 10^-10 = x^2/ 0.106-x

As HCN is weak, value of x is very small and 0.106-x ~ 0.106. Now the equation becomes

6.2 x 10^-10 = x^2/ 0.106

or, x = 0.810 *10^-5 M

Number of moles of H available from dissociation of HCN = 0.810 *10^-5 M * 0.1L = 0.810 *10^-6 moles

total moles of H+ in 100mL solution = 0.810 *10^-6 + 0.0106 M ~ 0.0106 moles

Molarity of the solution = 0.106 M

This means that as HCN is a weak acid, contribution from HCN is negligible.

pH =-log [H+] = -log[0.106] = 0.97

Initial pH = 0.97

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When NaOH is added, a buffer system will be generated (weak acid HCN and its salt NaCN). pH of this solution will be controlled by Hinderson Hasselbalch equation.

pH = pKa + log[salt/acid]

pKa = -logKa = -log[6.2 x 10^-10] = 9.2

[salt] = moles of base added = 0.075L * 0.106 M = 0.00795 moles

NaOH will react with initally present H+ to convert it to salt. Thus this amount has to be deducted.

moles of acid present = 0.0106 moles - 0.0075 = 0.0031 moles

pH = pKa + log[salt/acid]

     = 9.2 + log[0.00795/0.0031] = 9.6

pH of the solution after adding 75mL base = 9.6

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when 100mL base is added, all the acid wil be neutralize and [salt]= [acid]. So that log[salt/acid]= 0

pH = pKa = 9.2

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After the equivalence point,the salt formed will hydrolyze again.

CN- + H2O ==> OH- + HCN

Kb = [OH-][HCN]/[CN-]

Kb of HCN = Kw/ka = 1*10^-14/6.2*10^-10 = 0.16 *10-4

molarity of the NACN formed after adding 127mL NaOH = 100mL * 0.106 M/127+100 = 0.047 M

Kb = x^2/0.047-x

or, 0.16 *10-4 = x^2/0.047-x

or,(0.16*10^-4)(0.047-x) -x^2 = 0

or, 7.47*10^-3 -0.16*10^-4x -x^2 = 0

or, X = 0.086

[OH-] = 0.086 moles in 227 mL solution = 0.38 M

pOH = -log[OH-] = -log[0.38] = 0.42

pH = 14-pOH = 14-0.42 = 13.58 ~ 13.6

HCN H+ CN- initial 0.106 0 0 change -x +x +x equilibrium 0.106-x x x

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