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Chapter 10: Q1TY (page 218)

Find the pH of $0.002 {MK}^{+} {HP}^{-}$ with Equation $10 - 11$ .

Short Answer

Expert verified

The pH of given $K^{+} HP'$ solution is 4.28.

Step by step solution

Concept used.

Intermediate form of diprotic acid

$[H +] = \sqrt{\frac{K_{1} K_{2} I^{+} + K_{1} K_{w}}{K_{1} + F *}}$

Step2: Calculate the pH of 0.002MK+HP-.

The pH,

$[H +] = \sqrt{\frac{(4.7 \times 10^{- 3}) (1.8 \times 10^{- 10}) (0.002) + (4.7 \times 10^{- 3}) (1.0 \times 10^{- 14})}{(4.7 \times 10^{- 3}) + (0.002)}} pH = - \log [H^{+}] = 4.28$

The pH of $K^{+} HP'$ solution is 4.28.

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Find the pH of 0.002MK+HP-with Equation10-11.

Short Answer

Step by step solution

Concept used.

Step2: Calculate the pH of 0.002MK+HP-.

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Find the pH of $0.002 {MK}^{+} {HP}^{-}$ with Equation $10 - 11$ .