A 75.0-mL volumeof0 200mol L^-1 NH, (K_b 1.8 times 10^-5)is titrated with 0.500

Question

A 75.0-mL volumeof0 200mol L^-1 NH, (K_b 1.8 times 10^-5)is titrated with 0.500 mol L^-1 HNO_3. calculate the pH after the addition of 23.0ml of HNO_3 Express your answer numerically In case of a weak base strong acid titration the buffer solution accounts for the leveling of the titration curve in the buffer region between the start of the titration and the equivalence port A 52 0-mL volume of 0 350 mol L^-1 CH_3COOH (K_a = 1.8 times 10^-5) is titrated with 0 400 mol L^-1 NaOH Calculate the pH after the.addition of 19 0 mL of NaOH Express your answer numerically

Explanation / Answer

millimoles of CH3COOH = 52 x 0.350 = 18.2

millimoles of NaOH = 0.4 x 19 = 7.6

pKa = -logKa = -log (1.8 x 10^-5) = 4.74

CH3COOH + NaOH -------------------> CH3COONa + H2O

18.2 7.6 0 0

10.6 0 7.6 7.6

pH = pKa + log [salt /acid]

pH = 4.74 + log (7.6 / 10.6)

pH = 4.60

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