Question

calculate the hydroxide ion and hydroxyl ion concentration of solution containing 0.063 g of pure nitric acid per litre.​

Answer 1

Answer: $H^+$ = 0.001 M

$OH^-$ = $10^{-11}M$

Explanation:

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute  

$V_s$ = volume of solution in ml = 1000 ml   (1L=1000ml)

$moles of solute =\frac{\text {given mass}}{\text {molar mass}}=\frac{0.063g}{63g/mol}=0.001moles$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{0.001moles\times 1000}{1000ml}=0.001mole/L$

Therefore, the molality of solution will be 0.001 M.

$HNO_3\rightarrow H^++NO_3^-$

According to stoichiometry:

1 mole of $HNO_3$ gives 1 mole of $H^+$  

Thus 0.001 moles of $HNO_3$ gives 0.001 moles of $H^+$  

Thus $H^+$ = 0.001 M

$[H^+][OH^-]=10^{-14}$

$[OH^-]=\frac{10^{-14}}{[H^+]}$

$[OH^-]=\frac{10^{-14}}{[0.001]}=10^{-11}M$

Thus molarity of $H^+$ is 0.001 M and $[OH^-]$ in solution is $10^{-11}M$.