Question

Calculate the hydrogen and hydroxyl ion concentration in a solurion of 0.0315g of HNO3 in 500cc of water.

Answer 1

Explanation:

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Answer 2

The hydrogen and hydroxyl ion concentration in a solurion of 0.0315g of HNO3 in 500cc of water is $10^{-3}M$ and $10^{-11}M$ respectively.

Explanation:

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

where,

n= moles of solute  

$V_s$ = volume of solution in ml = 500 ml

$moles of solute =\frac{\text {given mass}}{\text {molar mass}}=\frac{0.0315g}{63g/mol}=5\times 10^{-4}moles$

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

$Molarity=\frac{5\times 10^{-4}moles\times 1000}{500ml}=0.001mole/L$

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 $H^+$

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

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

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

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