We define p value of a term say x as px -log x-Calculate pCl- value as x- 10-1 in a solution that is 2-00 - 10-3 M in NaCl and 5-4 - 10-4 M HCl-

The strong electrolytes, NaCl and HCl completely ionize in water. p_Na^+ = log [2 × 10^ 3]=3 log(2)=3 0.3010=2.6990 . p_H^+ = log[ 5.4 × ...

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Standard XII

Chemistry

Molarity

We define p v...

Question

We define p value of a term (say $x$ ) as $p_{x} = - l o g x$ .
Calculate $p_{(C l^{-})}$ value as $x \times 10^{- 1}$ in a solution that is $2.00 \times 10^{- 3}$ M in $N a C l$ and $5.4 \times 10^{- 4}$ M $H C l$ .

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Solution

The strong electrolytes, $N a C l$ and $H C l$ completely ionize in water.
$p_{N a^{+}} = - l o g [2 \times 10^{- 3}] = 3 - l o g (2) = 3 - 0.3010 = 2.6990$ .
$p_{H^{+}} = - l o g [5.4 \times 10^{- 4}] = 4 - l o g (5.4) = 4 - 0.7324 - 3.2676$ .
Total chloride ion concentration is $= 2 \times 10^{- 3} + 5.4 \times 10^{- 4} = 2.54 \times 10^{- 3} M$ .
Hence, $p_{(C l^{-})} = - l o g [2.54 \times 10^{- 3}] = 2.595 \approx 26 \times 10^{- 1}$ .

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We define p value of a term (say x) as px=−log x.Calculate p(Cl−) value as x×10−1 in a solution that is 2.00×10−3 M in NaCl and 5.4×10−4 M HCl.

We define p value of a term (say $x$ ) as $p_{x} = - l o g x$ .
Calculate $p_{(C l^{-})}$ value as $x \times 10^{- 1}$ in a solution that is $2.00 \times 10^{- 3}$ M in $N a C l$ and $5.4 \times 10^{- 4}$ M $H C l$ .