Calculate the pH of a buffer that is 0.225 M HC_2H_3O_2 and 0.162 M KC_2H_3O_2.

Question

Calculate the pH of a buffer that is 0.225 M HC_2H_3O_2 and 0.162 M KC_2H_3O_2. the K_a for HC_2H_3O_2 is 1.8 Times 10^-5. 4.589 9.11 4.74 9.26 4.62 Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH_4Cl with 100.0 mL of 0.20 M NH_3. The K_b for NH_3 is 1.8 Times 10^-5. 9.13 9.25 9.53 4.74 8398 Which of the following is TRUE? An effective buffer has a [base]/[acid] ratio in the range of 10 - 100. A buffer is most resistant to pH change when [acid] = [conjugate base] An effective buffer has a very small absolute concentrations of acid and conjugate base. A buffer can not be destroyed by adding too much strong base. It can only be destroyed by adding too much strong acid. None of the above are true. When titrating a monoprotic strong acid with a weak base at 25 degree C, the pH will be 7 at the equivalence point. pH will be greater than 7 at the equivalence point. titration will require more moles of the base than acid to reach the equivalence point. titration will require more moles of acid than base to reach the equivalence point. pH will be less than 7 at the equivalence point. A 100.0 mL sample of 0.18 M HClO_4 is titrated with 0.27 M LiOH. Determine the pH of the solution before the addition of any LiOH. 1.74 1.05 0.74 0.57 1.57 A 100.0 mL sample of 0.18 M HClO_4 is titrated with 0.27 M LiOh. Determine the pH of the solution after the addition of 30.0 mL of LiOh. 0.86 1.21

Explanation / Answer

VI

pH = pKa + log(A-/HA)

pH = 4.75 + log(0.162/0.225) = 4.6073

VII

pOH = pKb + log(NH4/NH3)

pOH = 4.75 + log(NH4/NH3)

mmol>

mmol of N# = 100*0.2

pOH = 4.75 + log(250*0.15/(100*0.2)) = 5.02300

pH = 14-5.02300 = 8.977

VIII

the most effectiv ebuffer is when

[Conjguate] = [Acid]

which is b

IX

if weak base --> ph will be acidic

HB+ + H2O <-> H3O+ + B

X

[H+] = -log(0.18)= 0.74

XI

mmol of acid = 100*0.18 = 18

mmol base = 0.27*30 = 8.1

18-8.1 = 9.9

[H+] = 9.9/(100+30) = 0.076153

pH = -log(0.076153) = 1.11

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