Calculate the ph of a solution resulting from the titration of 30.00 ml of 0.055 m hno3 with 40.00 ml of 0.026 m naoh

n(HNO3) =0.055 mol/L * 0.03000 L = 0.00165 mol HNO3

n(NaOH) = 0.026 mol/L * 0.04000 = 0.00104 mol NaOH

NaOH + HNO3 > NaNO3 + H2O

from equation 1 mol(NaOH) 1 mol(HNO3)

given 0.00104 mol (NaOH) 0.00165 mol(HNO3)

HNO3 is leftover 0.00165 - 0.00104 = 0.00061 mol HNO3

Volume of the solution is 30.00 + 40.00 = 70.00 ml = 0.07000 L

Concentration of HNO3 = 0.00061 mol HNO3/0.07000 L solution = 0.0087 mol/L

Concentration of H⁺ = 0.0087 mol/L, also because HNO3 ---> H⁺ + NO3⁻.

pH = - log{H⁺] = - log[0.0087] ≈ 2.06 ≈2.1

pH ≈ 2.1

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the answer is:  

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