Question

Calculate the pH when 64.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 × 10⁻⁹)

Answer 1

Answer:

The answer is "12.06"

Explanation:

Given:

$M(HBrO) = 0.3\ M\\\\V(HBrO) = 20 \ mL\\\\M(KOH) = 0.15 \ M\\\\V(KOH) = 64 \ mL$

$\to mol(HBrO) = M(HBrO) \times V(HBrO) = 0.3 M \times 20 mL = 6 \ mmol\\\\\to mol(KOH) = M(KOH) \times V(KOH)= 0.15 M \times 64 mL = 9.6 mmol$

  6 mmol of both will react

excess KOH remaining$= 3.15 \ mmol$

Volume$= 20 + 64 = 84 \ mL$

$[OH^{-}] = \frac{ 9.6 \ mmol}{84\ mL} = 0.01142\ M$

use:

$pOH = -\log [OH^-]$

        $= -\log (1.142\times 10^{-2})\\\\= 1.94$

use:

$PH = 14 - pOH$

       $= 14 - 1.94\\\\= 12.06$