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Standard XII

Chemistry

Solubility Product

A solution of...

Question

A solution of $H_{2} C_{2} O_{4}$ requires 200ml of NaOH for complete neutralization. The pH of the solution when 120ml of NaOH is added is: $($ Given $: K a_{1} = 10^{- 4}$ and $K a_{2} = 10^{- 7}$ for $H_{2} C_{2} O_{4}) [log 2 = 0.30]$

3.4

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6.4

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Solution

The correct option is A 6.4
One mole of $H_{2} C_{2} O_{4}$ requires 2 moles of NaOH for complete neutralization. Thus, when 120 ml is used:
$H_{2} C_{2} O_{4} + N a O H \to N a H C_{2} O_{4} + H_{2} O$
100x 200x 0
0 100x 100x

Again,
$N a H C_{2} O_{4} + N a O H \to N a_{2} C_{2} O_{4}$
100x 20x 0
80x 0 20x
$∴ p H = p K a_{2} + log \frac{20 x}{80 x} = - l o g (10^{- 7}) + l o g \frac{1}{4}$

$∴ p H = 6.4$

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A solution of H2C2O4 requires 200ml of NaOH for complete neutralization. The pH of the solution when 120ml of NaOH is added is: (Given :Ka1=10−4 and Ka2=10−7 for H2C2O4)[log2=0.30]

The correct option is A 6.4One mole of H2C2O4 requires 2 moles of NaOH for complete neutralization. Thus, when 120 ml is used:H2C2O4+NaOH→NaHC2O4+H2O100x 200x 0 0 100x 100x Again,NaHC2O4+NaOH→Na2C2O4 100x 20x 0 80x 0 20x∴pH=pKa2+log20x80x=−log(10−7)+log14 ∴pH=6.4

A solution of $H_{2} C_{2} O_{4}$ requires 200ml of NaOH for complete neutralization. The pH of the solution when 120ml of NaOH is added is: $($ Given $: K a_{1} = 10^{- 4}$ and $K a_{2} = 10^{- 7}$ for $H_{2} C_{2} O_{4}) [log 2 = 0.30]$