Calculate pH

Chemistry Level 2

Calculate the pH of a 5.6 L solution made by dissolving 0.46 mol of NaH ( s ) \text{NaH}(s) in water.

1.09 2.09 11.91 12.91

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1 solution

Sowmitra Das
May 12, 2014

N a H NaH reacts with water as follows:
N a H + H 2 O N a O H + H 2 NaH+H_2O \rightarrow NaOH+H_2

So, one mole of N a H NaH produces 1 1 mole of N a O H NaOH .
Since, N a O H NaOH ionizes completely, we have, [ O H ] = [ N a O H ] [ N a H ] = 0.46 5.6 [OH^{-}]=[NaOH]\equiv[NaH]=\frac{0.46}{5.6}

p O H = log [ O H ] = log 0.46 5.6 \therefore pOH=-\log{[OH^-]}=-\log{\frac{0.46}{5.6}}
p H = 14 p O H = 14 + log 0.46 5.6 12.91 \Rightarrow pH=14-pOH=14+\log{\frac{0.46}{5.6}}\approx \boxed{12.91}

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