Question

# A 20.0 mL of 0.20 M HNO3 was titrated with 10.0 mL of 0.20 M NaOH....

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.

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

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