0.63 g of `HNO_(3)` dissolved in 100 ml of solution. Calculate its pH value.

Explanation

pH of 0.63g `HNO_(3)` in 10 litres of solution.
No. of moles of `HNO_(3) = (0.63)/(63) = 0.01` mol
Molarity (M) `= (0.01)/(10) = 10^(-3)`
Now, `[H^(+)] = 10^(-3)`
`pH = -log [H^(+)] = -log [10^(-3)] = -(-3 log 10) = 3`.