Question

A solution has been prepared by dissolving 0.063g of hno3 in 1000ml of it.Hence molarity of [oh]ion in solution

Answer 1

Answer: $10^{-11}M$.

Explanation:

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

where,

n= moles of solute  

$V_s$ = volume of solution in ml = 1000 ml

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

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 $H^+$

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

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

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

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