Question

What is the pH of 10.00 mLs of a 0.100 Molar formic acid solution (ka = 1.9 x 10^-4) after the following additions: A. 0.00 mLs of 0.100 M lithium hydroxide, B. 5.00 mLs of 0.100 M lithium hydroxide, C. 10.00 mLs of 0.100 M lithium hydroxide, D. 15.00 mLs of 0.100 M lithium hydroxide

Answer 1

pH of bufer solution = pKa + log ([salt]/[acid])

buffer solution : When weak acid like formic acid added to the strong base like LiOH, buffer solution forms.

pKa of acid = - log(Ka) = -log (1.9 x 10^-4) = 4 - log1.9 = 4-0.2787 = 3.72

Formic acid +  lithium hydroxide -----------> Lithium formate + water

A.

volume of acid = 10 mL; concentration = 0.100 M

Volume of acid = 0 mL; concentration = 0.100 M

There is no salt formation because there is no base addition (vol is zero)

So the acid solution pH = -logH⁺

HCOOH <-----------> HCOO^- + H⁺

concentration c-x x x

Ka = [HCOO^- ] [H⁺] / HCOOH

Ka = X ²/ c-X

(C - x)K_a = x²

x² + K_ax - CK_a = 0

x = [-K_a + (K_a² + 4CK_a)^½]/2

K_a = 1.9 x 10^-4 and C = 0.1 M

X = [-1.9 x 10^-4 + {(1.9 x 10^-4)² + 4 *0.1*1.9 x 10^-4}^1/2 / 2 = [-1.9*10^-4 + 2.09*10^-6]/2 = 0.19*10^-2 /2 = 0.095*10^-2

[H+] = X = 95*10^-5

pH = -log(95*10^-5) = 5 - log95 = 5 - 1.98 = 3

B. 5.00 mLs of 0.100 M lithium hydroxide,

Ans.

Formic acid +  lithium hydroxide -----------> Lithium formate + water

volume of acid = 10 mL; concentration = 0.100 M ; mmoles = 0.1 * 10 /1000 = 1 *10^-3 mol

Volume of base = 5 mL; concentration = 0.100 M; mmoles = 0.1 * 5/1000 =0.5 *10^-3 mol

(0.5*10^-3 mol)of base reacts with (1 *10^-3 mol) of acid and gives (0.5 *10^-3 mol) of salt and the total volume is 15 mL.

So the M =0.5*10^-3/ 15 = 0.033* 10^-3 M

concentration of remaiing acid = 05 mL of Formic acid will be free because one eqiv of base reacts with one eqiv of acid. So 5 mL of acid still remains.  

moles = 5 * 0.1 /1000 = 0.5 *10^-3 ; total volume = 15 mL

conc of acid =0.5 *10^-3 / 15 = 0.033 *10^-3

pKa = 3.72

pH of bufer solution = pKa + log ([salt]/[acid])

pH = 3.72 + log (0.033 *10^-3) / (0.033 *10^-3) = 3.72+ 0 = 3.72

C. 10.00 mLs of 0.100 M lithium hydroxide

Formic acid +  lithium hydroxide -----------> Lithium formate + water

volume of acid = 10 mL; concentration = 0.100 M ; mmoles = 0.1 * 10 /1000 = 1 *10^-3 mol

Volume of acid = 10 mL; concentration = 0.100 M; mmoles = 0.1 * 10/1000 = 1 *10^-3 mol

(1*10^-3 mol)of base reacts with (1 *10^-3 mol) of acid and gives (1 *10^-3 mol) of salt and the total volume is 20 mL.

There is no excess of acid or base present. The salt only present. So the solution pH will be neutral .

pH will be near 7

D. 15.00 mLs of 0.100 M lithium hydroxide

Formic acid +  lithium hydroxide -----------> Lithium formate + water

volume of acid = 10 mL; concentration = 0.100 M ; mmoles = 0.1 * 10 /1000 = 1 *10^-3 mol

Volume of acid = 15 mL; concentration = 0.100 M; mmoles = 0.1 * 15/1000 = 1.5 *10^-3 mol

(1*10^-3 mol)of base reacts with (1 *10^-3 mol) of acid and gives (1 *10^-3 mol) of salt and the total volume is 25 mL.

There is no acid left but 5 mL of excess base is present. So the concentation of OH will decide the pH

moles = 5 * 0.1 /1000 = 0.5 *10^-3 ; total volume = 25 mL

conc of OH =0.5 *10^-3 / 25 = 2 *10^-5

pOH = -log[OH] = - log 0.02 *10^-3 = 5 - 0.3010 = 4.7

pH = 14 -pOH = 14 - 4.7 = 9.3

pH o the solution is 9.3