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Molarity

50 mL of 0....

Question

$50$ mL of $0.1 M N a O H$ is added to 60 mL of 0.15 M $H_{3} P O_{4}$ solution . The $p H$ of the mixture would be about:
$(K_{1}, K_{2}$ and $K_{3}$ for $H_{3} P O_{4}$ are $10^{- 3}, 10^{- 8}$ and $10^{- 13}$ respectively)
$(l o g 2 = 0.3)$

3.1

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Solution

The correct option is A 3.1
50 mL of 0.1 M NaOH $- \frac{50}{1000} \times 0.1 = 0.005$ moles
60 mL of 0.15 M H3PO4 $- \frac{60}{1000} \times 0.15 = 0.009$ moles
0.005 moles of NaOH reacts with 0.005 moles of $H_{3} P O_{4}$ to form 0.005 moles of $N a H_{2} P O_{4}$
$0.009 - 0.005 = 0.004$ moles of $H_{3} P O_{4}$ remains
$p H = - l o g K_{a} + l o g \frac{[N a H_{2} P O_{4}]}{[H_{3} P O_{4}]}$
$p H = - l o g 10^{- 3} + l o g \frac{0.005}{0.004} = 3.1$

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Similar questions

50 m L

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is added to

60 m L

0.15 M H_{3} P O_{4}

solution (

K_{1}, K_{2}

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p H

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G i v e n K_{a 1} = 10^{- 3}, K_{a 2} = 10^{- 8}, K_{a 3} = 10^{- 13}

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50 mL of 0.1 M NaOH is added to 60 mL of 0.15 M H3PO4 solution . The pH of the mixture would be about: (K1,K2 and K3 for H3PO4 are 10−3,10−8 and 10−13 respectively)(log 2=0.3)

$50$ mL of $0.1 M N a O H$ is added to 60 mL of 0.15 M $H_{3} P O_{4}$ solution . The $p H$ of the mixture would be about:
$(K_{1}, K_{2}$ and $K_{3}$ for $H_{3} P O_{4}$ are $10^{- 3}, 10^{- 8}$ and $10^{- 13}$ respectively)
$(l o g 2 = 0.3)$