Question

Calculate the pH change when 10. mL of 3.0 M HCl are added to 500. mL of the following

A ) pure water

B) aqueous solution of 3.0g of formic acid

C) aqueous solution of 4.0g potassium formate

D) aqueous solution containing 3.0 g of formic acid and 4.0 g of potassium formate

Answer 1

The pH of solution will be

a) pure water (500mL) so the HCl will get diluted and the H+ concentration will decrease

M1V1 = M2V2

3 X 10 = M2 X 510

M2 = 0.059 M

[H+] = 0.059

pH = -log[H+] = 1.23

b) the pH will be due to both formic acid and HCl

H+ concentration due to HCl = 0.059

H+ concentration due to formic acid will be calcualted by calculating the molarity of solution

molecular weight = 46g / mole

moles of formic acid = Mass / Molecular weight = 3 / 46 = 0.065 moles

Molarity = Moles / Volume in litres = 0.065 / 0.510 = 0.127

For weak acids

[H+] = (Ka X concentration )^1/2

Ka for fromic acid =.00018

[H+] = (.00018 X 0.127)^1/2 = 0.0048

Total concentration of [H+] = 0.0048 + 0.059 = 0.0638

pH = -log[H+] = 1.195

c) the potassium formate will react with HCl as

the moles of potassium formate present = mass / mol wt = 4 / 84.12 = 0.0475

Moles of Hcl added = 3 X 0.001 = 0.03 moles

So it will react with potassium formate to from formic acid

HCl + HCOONa --> HCOOH + NaCl

so it will result into formation of buffer

moles of salt present = Moles of salt initially present - moles of acid added = 0.0475 - 0.03 = 0.0175

Moles of weak acid present = moles of Hcl added = 0.03

ph of buffer is

pH =pKa + log [salt] / [acid]

pKa = -logKa = 3.74

pH = 3.74 + log [0.0175 / 0.03] = 3.51

d) already it is a buffer of formic acid and potassium formate with moles

Formic acid = 0.065

Potassium formate = 0.0475

Now again the moles of acid added = 0.03 moles

It will react with 0.03 moles of salt to form more of 0.03 moles of weak acid

moles of weak acid = 0.03 + 0.065 = 0.095

Moles of salt = 0.0175

pH = pKa + log [salt] / [acid]

pH = 3.74 + log 0.0175 / 0.095] = 3.005