The pH at the equivalent point for the titration of 0.10M K H _ 2 B O _ 3 with 0.1MHCl is ( K _ a of H _ 3 BO _ 3 =12 .8 × 10 ^ - 10 ) Report your answer by rounding it up to nearest whole number.

Question

Quantrex Academy · Accepted Answer

Correct answer: 5

First there will be an acid-base reaction between

K{H}_{2}B{O}_{3} and HCl

.

The pH at the equivalent point for the titration of 0.10M K H _ 2 B O _ 3 with 0.1MHCl is ( K _ a of H _ 3 BO _ 3 =12

conc. will become half after reaction of an equal volume of

{KH}_{2}{BO}_{3} and HCl

So

\dfrac{0.1}{5} = 0.05

Now,

\left[{H}^{+}\right] = \sqrt{{K}_{a}C} = \sqrt{12.8 \times {10}^{- 10} \times 0.05} = 8 \times 1{0}^{- 6}

pH = - \log \left(8 \times {10}^{- 6}\right) \approx 5.0

The pH at the equivalent point for the titration of 0.10M K H _ 2 B...

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