Question

calculate the H3O+ and OH- Ion concentration at 25° c in
(i) 0.02 N HCL solution and
(ii) 0.005 N NaOH solution

Answer 1

am I right????????????

Answer 2

$[H_3O^+]$ and $[OH^-]$ are 0.02 M and $5\times 10^{-13}M$ for 0.02 N HCl.

$[H_3O^+]$ and $[OH^-]$ are $2\times 10^{-12}M$ and 0.005 M for 0.005 N NaOH.

Explanation:

Normailty is defined as number of gram equivalents present per litre of the solution.

$Normality=Molarity\times n$

n= basicity or acidity = number of replaceable hydrogen or hydroxide ions

(i) 0.02 N HCl , $Molarity=\frac{Normality}{n}=\frac{0.02}{1}=0.02M$

1 mole of HCl gives = 1 mole of $H_3O^+$

Thus 0.02 mole of HCl gives =$\frac{1}{1}\times 0.02=0.02$ moles of $H_3O^+$

$[OH^-]=\frac{10^{-14}}{H^+}=\frac{10^{-14}}{0.02}=5\times 10^{-13}M$

(ii) 0.005 N NaOH, $Molarity=\frac{Normality}{n}=\frac{0.005}{1}=0.005M$

1 mole of NaOH gives = 1 mole of $OH^-$

Thus 0.005 mole of NaOH gives =$\frac{1}{1}\times 0.005=0.005$ moles of $OH^-$

$[H_3O^+]=\frac{10^{-14}}{OH^-}=\frac{10^{-14}}{0.005}=2\times 10^{-12}M$

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