Question

How does pH of 0.5L of pure water change upon addition of 0.5ml of 1M HCL?

0.5 ml of 1 M NaOH was added to 0.5L of a 100mM buffer solution at pH=7.0. Calculate how the pH value is affected by this addition if a pKa value of the buffer is 7.47.

Answer 1

1) The pH of pure water at room temperature = 7,

According to the definition of pH, -Log[H⁺] = 7

i.e. [H⁺]_{in pure water} = 10^-7 M (= M₁)

The volume of pure water = 0.5 L = V₁

The concentration of added HCl, i.e. [H⁺]_{in HCl} = 1 M (= M ₂)

The volume of added HCl = 0.5 mL, i.e. 0.5*10^-3 L = V₂

The molarity of resulting solution = (M₁V₁ + M₂V₂) / (V₁ + V₂)

= (10^-7*0.5 + 1*0.5*10^-3) / (0.5 + 0.5*10^-3)

~ 0.5*10^-3 / 0.5 (since 0.5*10^-3 >> 0.5*10^-7 and 0.5 >> 0.5*10^-3)

~ 10^-3 M

Therefore, the pH of resulting solution = -Log(10^-3), i.e. 3

Hence, the pH of pure water decreases from 7 to 3 upon addition of 0.5 mL of 1M HCl.

2) According to Henderson-Hasselbalch equation, we can write as shown below.

pH = pKa + Log{[HA]/[A^-]}

Here, a solution of NaOH (i.e. basic solution) is added to the given buffer solution at pH = 7 (i.e. neutral solution)

Therefore, the pH of the resulting solution increases. (since the pH of strong base > 7)

i.e. The pH of resultiong solution > 7.