9/ A Student is asked to prepare a buffer solution at pH = 3.6, which one of the
ID: 854729 • Letter: 9
Question
9/ A Student is asked to prepare a buffer solution at pH = 3.6, which one of the following weak acids should be used? (a) NH4OH (Kb=1.8x10-5) (b) HA (Ka=2.7 x10-4) (c) HB (Ka=1.5x10-9) (d) HC (Ka=5.3x10-13)
10/ Which one of the following statement is FALSE? In the equilibrium Mg(OH)2 (s)=> Mg2+(aq) + 2OH-(aq) (a) the reaction moves forward when adding in acid (b) The reaction moves backward when adding in solid NaCl (c) The reaction moves backward when adding in solid MgCl2 (d) The reaction moves forward when adding in water
11. Exactly 100mL of 0.10 M HNO2 are titrated with a 0.10 M NAOH solution. Calculate the pH for (a) the initial solution, (b) the point at which 80 mL of the base has been added, (c) the equivalence point, (d) the point at which 105 mL of the base has been added. Ka=4.5x10-4
Explanation / Answer
11)
This is for 0.11 M HNO2 are titrated with a 0.11 M NAOH solution. please substitute values and do.
HNO2 <----> H+ + NO2-
Ka = 4.5 x 10^-4 = x^2 / 0.11 -x
x = [H+] = 0.00704 M
pH = 2.15
Moles HNO2 = 100 x 0.11 / 1000 = 0.011
Moles NaOH = 80 x 0.11 / 1000 = 0.0088
0.011 mole HNO2 + 0.0088 mole NaOH >> 0.0022 mole HNO2 + 0.0088 mole NO2-
Total volume = 200 mL = 0.200 L
[HNO2] = 0.0022 / 0.200 = 0.011 M
[NO2-] = 0.0088 / 0.200 = 0.044 M
pKa = - log Ka = 3.35
pH = pKa + log 0.044 / 0.011 = 3.95
Moles HNO2 = moles NaOH = 0.011
Total volume = 0.200 L
[NO2-] = 0.011 / 0.200 = 0.055 M
Kh = 1 x 10^-14 / 4.5 x 10^-4 = 2.22 x 10^-11
NO2- + H2O <---> HNO2 + OH-
2.22 x 10^-11 = x^2 / 0.055 -x
x = 1.11 x 10^-6 M
pOH = 5.96
pH = 8.04
Moles NaOH = 105 x 0.11 / 1000 =0.01155
Moles HNO2 = 100 x 0.11 / 1000 = 0.011
Moles OH- in excess =0.00055
Volume = 205 mL = 0.205 L
[OH-] = 0.00055 / 0.205 = 0.00268 M
pOH = 2.57
pH = 11.43
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