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Standard X

Chemistry

Mole Concept

How many mole...

Question

How many mole of $H C N$ will be required to prepare one litre buffer solution of $p H = 10.4$ using $0.01$ mol $N a C N$ ?
(Given: $K_{a} (H C N) = 4 \times 10^{- 10}$ )

2 \times 10^{- 2}

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9 \times 10^{- 4}

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4 \times 10^{- 4}

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9 \times 10^{- 3}

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Solution

The correct option is B $9 \times 10^{- 4}$ M
$p^{H} = p^{K a} + l o g (\frac{s}{a})$
$p^{H} = 10.4$
$p^{K a} = - l o g (K a) = 9.397$
$10.4 = 9.397 + l o g (\frac{0.01}{a})$
$1.002 D = l o g (\frac{0.01}{a})$
$10.04 θ = \frac{0.01}{a}$
$a = \frac{0.01}{10.048} \Rightarrow a = 9.95 \times 10^{- 4} M$
Correct answer

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Similar questions

How many gram moles of $H C l$ will be required to prepare one litre of a buffer solution (containing $N a C N$ and $H C N$ ) of $p H 8.5$ using $0.01 g r a m$ formula weight of $N a C N$ ?

$K_{a} (H C N)$ = $4.1 \times 10^{- 10}$

Q. How many gram moles of

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K_{a} (H C N) = 4.1 \times 10^{- 10}

Q. A buffer of pH

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0.02

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(K_{a} H C N = 6.2 \times 10^{- 10})

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How many mole of HCN will be required to prepare one litre buffer solution of pH=10.4 using 0.01 mol NaCN? (Given: Ka(HCN)=4×10−10)

The correct option is B 9×10−4MpH=pKa+log(sa)pH=10.4pKa=−log(Ka)=9.39710.4=9.397+log(0.01a)1.002D=log(0.01a)10.04θ=0.01aa=0.0110.048⇒a=9.95×10−4MCorrect answer

How many mole of $H C N$ will be required to prepare one litre buffer solution of $p H = 10.4$ using $0.01$ mol $N a C N$ ?
(Given: $K_{a} (H C N) = 4 \times 10^{- 10}$ )