Question

Calculate the pH of a solution made by adding 25.0 g of sodium formate, NaHCOO, to 400 ml of 0.64 M formic acid, нсоон. Answer: What is the pH of pure water at 25°C? Answer: When a small amount of acid is added to a non-buffered solution, there is a large change in pH. Calculate the pH when 22.2 ml of 0.0020 M HCl is added to 100.0 mL of pure water. Comment and hint in the general feedback Answer:

Answer 1

Molar mass of sodium formate

HCOONa = 1.0079 + 12.01 + ( 2 $\times$ 16.00 ) + 22.99 = 68.01 g/mol

We know that, No. of moles = Mass / Molar mass

$\therefore$ No. of moles of HCOONa = 25.0 g / 68.01 g/ mol = 0.3676 mol

Volume of solution = 400.0 ml = 0.4000 L

[HCOONa] = No. of moles of HCOONa / volume of solution in L

= 0.3676 mol / 0.4000 L

= 0.919 M

pH of buffer solution is calculated by using Henderson's equation.

pH = pKa + log [Salt] / [Acid]

pH = 3.744 + log [HCOONa] / [HCOOH]

pH = 3.744 + log 0.919 / 0.64

pH = 3.744 + 0.157

pH = 3.901

PART 2

pH of pure water.

Consider auto ionization of water, 2 H₂O $\rightleftharpoons$ H₃O ⁺ + OH ^-

For above reaction, equilibrium constant is K w = [ H₃O ⁺] [OH ^- ]

At 25 ⁰ C , value of K w = 1 $\times$ 10 ^-14

$\therefore$ K w = [ H₃O ⁺] [OH ^- ] = 1 $\times$ 10 ^-14

At equilibrium , [ H₃O ⁺] = [OH ^- ]

$\therefore$ [ H₃O ⁺] [ H₃O ⁺] = 1 $\times$ 10 ^-14

[ H₃O ⁺] ² = 1 $\times$ 10 ^-14

[ H₃O ⁺] = 1 $\times$ 10 ^-07 M

We have, pH = -log [ H₃O ⁺] = - log  1 $\times$ 10 ^-07 = 7.0

PART 3

When HCl is added to water , solution becomes dilute. pH will depend on concentration of diluted HCl solution.

We can use dilution formula to find out concentration of diluted solution.

We have formula, M Stock $\times$ V stock = M dilute $\times$ V dilute

M dilute = M Stock $\times$ V stock / V dilute

Here volume of dilute solution = 100.0 + 22.2 = 122.2 ml

Molarity of stock HCl solution = 0.0020 M

Volume of stock HCl solution = 22.2 ml

$\therefore$ M dilute = 0.0020 M $\times$ 22.2 ml / 122.2 ml = 0.000363 M

We have, pH = - log [ H₃O ⁺] = - log 0.000363 = 3.44

ANSWER : pH = 3.44