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Buffer Solutions

A buffer solu...

Question

A buffer solution is prepared in which the concentration of $N H_{3}$ is $0.30 M$ and the concentration of $N H_{4}^{+}$ is $0.20 M$ . If the equilibrium constant, $K_{b}$ for $N H_{3}$ equals $1.8 \times 10^{- 5}$ , what is the pH of this solution? $(l o g 2.7 = 0.433)$

9.08

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9.43

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11.72

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8.73

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Solution

The correct option is A 9.43
$[N H_{3}] = 0.3 M; [N H_{4}^{+}] = 0.2 M$

$k_{b} = 1.8 \times 10^{- 5}$

$p O H = P k_{b} + l o g \frac{[s a l t]}{[b a s e]} = 4.74 + l o g (\frac{0.2}{0.3})$

$= 4.74 + 0.3010 - 0.4771 = 4.56$

$p H = 14 - 4.56 = 9.436$

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Similar questions

N H_{4}^{+}

0.02 M N H_{3}

0.01 M K O H

(K_{b}

N H_{3} = 1.8 \times 10^{- 5} M)

N H_{3} a n d N H_{4}^{+} C l^{-}

m o l L^{- 1} .

p K_{b}

N H_{3} = 4.7, l o g 2 = 0.3

M n (O H)_{2}

k_{s p}

\times 10^{- 14}

{N H}_{4}^{+}

N H_{3}

k_{b} = 1.8 \times 10^{- 5}

N H_{3} a n d N H_{4}^{+} C l^{-}

m o l L^{- 1} .

p K_{b}

N H_{3} = 4.7, l o g 2 = 0.3

[{N H}_{4}^{+}]

{N H}_{3}

K_{b} = {1.8 \times 10}^{- 5}

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