Question

A 20.0 mL of 0.20 M HNO₃ was titrated with 10.0 mL of 0.20 M NaOH. Determine the pH of the solution and record below with correct significant figures.

Answer 1

Here we are having 20 ml of 0.20 M HNO₃ .

So Number of moles of H⁺ = Volume * molarity = 20* 10^-3 l * 0.20 mol l^-1 = 0.004 moles

Similarly we have 10.0 mL of 0.20 M NaOH .

So Number of moles of OH^- =  Volume * molarity = 10* 10^-3 l * 0.20 mol l^-1 = 0.002 moles.

Each mole of OH^- reacts with one mole of H⁺ . So all 0.002 moles of OH^- will react with corresponding 0.002 moles of H⁺.

Number of moles of  H⁺ left = 0.004 mol - 0.002 mol = 0.002 mol

Meanwhile Total volume of solution has become 40 ml since we have mixed 20 ml of HNO₃ and 20 ml of NaOH.

Hence , Concentration of H⁺ = Moles of H⁺ / Volume of solution ( in L ) = 0.002 mol / 0.040 L = 0.05 M

pH of solution = -log [ H⁺ ] = - log(0.05) = 1.301