Exactly 150mL of 0.13M HNO2 is titrated with 100mL of 0.13 M NaOH. What is the p

Question

Exactly 150mL of 0.13M HNO2 is titrated with 100mL of 0.13 M NaOH. What is the pH at the endpoint.

Explanation / Answer

moles HNO2 = 0.020 L x 0.27 M=0.0054 moles KOH needed = 0.0054 Volume KOH = 0.0054 mol/ 0.15 M=0.036 L total volume = 0.036 + 0.020 = 0.056 L HNO2 + OH- => NO2- + H2O moles NO2- produced = 0.0054 [NO2-]= 0.0054/ 0.056L=0.096 M NO2- + H2O HNO2 + OH- initial concentration 0.096 change -x .. . . . . .. . . . . . . .. . . +x. . .. . . +x at equilibrium 0.096-x. .. . . . . .. . . .. .x. . .. . . .. x Kb ( also called hydrolysis constant )= Kw/Ka = 1 x 10^-14/ 6 x 10^-4=1.7 x 10^-11 1.7 x 10^-11 = x^2 / 0.096-x x = [OH-]= 1.3 x 10^-6 M pOH = 5.9 pH = 14 - 5.9=8.1

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