Learning Goal: To calculate the pH at the equivalence point for various types of

Question

Learning Goal: To calculate the pH at the equivalence point for various types of titrations. A84.0mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 42.0mL of KOH. Express the pH numerically. Consider the titration of 50.0 mL of 0.20 M NH3 (Kb = 1.8 Times 10 -5) with 0.20 M HNO3 Calculate the pH after addition of 50.0 mL of the titrant. Express the pH numerically. A30.0-mL volume of 0.50 M CH3COOH (Ka = 1.8 Times 10-5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH. Express the pH numerically.

Explanation / Answer

Reaction NH3 + H2O --> NH4+ + OH-

initially 0.2 0 0

finally 0.2-x x x

ka = 1.8*10^-5= [NH4+]*[OH-]/[NH3]

1.8*10^-5 = x^2/0.2-x

x=1.88*10^-3

[OH-]= 1.88*10^-3M

now 0.2M HNO3 is added 50ml

[H+]= (0.2*50-1.88*10^-3*50)/(50+50)

[H+] = 0.09906

pH = -log10[H+]

pH= 1.004

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