Question

pH of aqueous solution of potassium hydroxide at 298 K temperature is 11.65. The initial volume of this solution is made 6 times by addition of water. What will be the pH of the diluted solution? Solve the given problem.

Answer 1

Given, pH of KOH = 11.65
we know, pH + pOH = 14
so, pOH of KOH = 14 - 11.65 = 2.35
now use formula, pOH = -log₁₀[OH-]
2.35 = -log₁₀[OH-]
log₁₀[OH-] = - 2.35 = 3 - 2.35 - 3
log₁₀[OH-] = 0.65 - 3 = -3 + 0.65
log₁₀[OH-] = $\overline{3}$ + 0.65
now taking antilog ,
[OH-] = antilog ($\overline{3}$ + 0.65)
= 4.46 × 10^-3

now, according to question, if volume is made 6 times then the concentration will decrease 6 times.
So the new concentration is = 4.46× 10^-3/ 6
= 7.44 × 10^-4
So new pOH = -log₁₀[OH-]
= -log₁₀(7.44 × 10^-4)
= 4-log₁₀[7.44]
= 4- 0.871
= 3.129
hence, pOH = 3.129
Therefore, new pH is 14 – 3.129 = 10.871