Part A. Calculate the pH of a buffer solution that is 0.250 M in HCN and 0.168 M
ID: 960634 • Letter: P
Question
Part A. Calculate the pH of a buffer solution that is 0.250 M in HCN and 0.168 M in KCN. For HCN, Ka = 4.9×1010 (pKa = 9.31).
Part B. Calculate the pH of a buffer solution that is 0.240 M in HC2H3O2 and 0.190 M in NaC2H3O2. (Ka for HC2H3O2 is 1.8×105.)
Part C. Consider a buffer solution that is 0.50 M in NH3 and 0.20 M in NH4Cl. For ammonia, pKb=4.75. Calculate the pH of 1.0 L of the original buffer, upon addition of 0.150 mol of solid NaOH.
Part D. A 1.0-L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×105. Calculate the pH of the solution, upon addition of 0.055 mol of NaOH to the original buffer.
Explanation / Answer
a.
The Henderson -Hasselbalch equation is: pH=pKa+log([A-]/[HA])
where [HA] is the molarity of the acid and [A-] is the molarity of its conjugate base.
pH=9.31+log(0.170/0.250)
=9.14
b.
pH = pKa + log 0.190/0.250
Ka = 1.8 x 10^-5
pKa = -log (1.8 x 10^-5) = 4.74
pH = 4.6
c.
pH = 14 - pKb + log([NH3 moles]/[NH4Cl moles])
INITIAL CONDITION
NH3 moles = 0.5 * 1.0 = 0.5 mol
NH4Cl moles = 0.2 * 1.00 = 0.2 mol
pH = 14.00 - 4.75 + log([0.50]/[0.2]) = 8.852
and this is the starting condition.
You add a small amount of Sodium Hydroxide. Chemical Buffer tend to slow-down the pH-influences exercited by Acid/Bases.
NH4Cl(aq) + NaOH(aq) <---> NH3(aq) + NaCl(aq) + H2O(aq)
hence NH3 IS FORMING while NH4Cl IS SINKING.
NH3 moles = 0.5 + 0.15 = 0.35 mol
NH4Cl moles = 0.2 - 0.15 = 0.05 mol
and the equation will be
pH = 14.00 - 4.75 + log([0.35]/[0.05]) = 8.4049
d.
moles of NaOH : .055 + .100 = .155moles
moles of acid : .1 - .055 = 0.045
pH = pKa + log [A]/[HA]
pH = 4.75 + log [.155]/[.045]
pH = 5.287
pH=4.87600