The base imidazole has $K_{b}$ of $1.0 \times 10^{- 7}$ . What volumes of 0.02 M HCl and 0.02 M imidazole should be mixed to make 150 mL of a buffer of pH 7 ?

100 ml and 50 ml

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50 mL and 100 mL

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75 ml and 75 ml

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60 ml and 90 ml

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Solution

The correct option is B
50 mL and 100 mL

Let x ml of the acid be used. Volume of base mixed = 150 - xml
Volume of base left = 150 - x - x = 150 - 2x
pOH = 14 - pH = 14 - 7 = 7
$= p K_{b} + l o g \frac{[S a l t]}{[B a s e]} = - l o g 1.0 \times 10^{- 7} + l o g \frac{x}{150 - 2 x}; l o g \frac{x}{150 - 2 x} = 0 = l o g 1 o r, \frac{x}{150 - 2 x} = 1 \Rightarrow x = 50 m l$

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The base imidazole has Kb of 1.0×10−7. What volumes of 0.02 M HCl and 0.02 M imidazole should be mixed to make 150 mL of a buffer of pH 7 ?

The correct option is B 50 mL and 100 mL Let x ml of the acid be used. Volume of base mixed = 150 - xml Volume of base left = 150 - x - x = 150 - 2x pOH = 14 - pH = 14 - 7 = 7 =pKb+log[Salt][Base]=−log 1.0×10−7+logx150−2x;logx150−2x=0=log 1or,x150−2x=1⇒x=50ml

The base imidazole has $K_{b}$ of $1.0 \times 10^{- 7}$ . What volumes of 0.02 M HCl and 0.02 M imidazole should be mixed to make 150 mL of a buffer of pH 7 ?