For each of the following solutions, calculate the initial pH and the final pH a
ID: 957513 • Letter: F
Question
For each of the following solutions, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH.
Part A
For 290.0 mL of pure water, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH.
Express your answers using two decimal places separated by a comma.
Answer: pHinitial, pHfinal = 7.00,12.84
Part B
For 290.0 mL of a buffer solution that is 0.200 M in HCHO2 and 0.310 M in KCHO2, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH(Ka=1.8104).
Express your answers using two decimal places separated by a comma.
Answer: pHinitial, pHfinal = 3.94,4.21
Part C
For 290.0 mL of a buffer solution that is 0.2975 M in CH3CH2NH2 and 0.2675 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH(Kb=5.6104).
Express your answers using two decimal places separated by a comma.
I CAN'T FIGURE THIS ONE OUT!
Explanation / Answer
part A
for pure water pH = 7
concentration of solution = 0.02/0.29 = 0.07 M
pOH = -log(OH-)
= -LOG0.07 = 1.155
pH = 14-1.155 = 12.845
part B
No of mol of HCHO2 = 0.2*0.29 = 0.058 mol
No of mol of KCHO2 = 0.31*0.29 = 0.09 mol
pH = pka + log(salt/acid)
pka = -logKa = -log(1.8*10^-4) = 3.74
= 3.74+log(0.09/0.058)
= 3.93
after addition of NaOH
= 3.74+log((0.09+0.02)/(0.058-0.02))
= 4.21
part C
No of mol of CH3CH2NH2 = 0.2975*0.29 = 0.0863 mol
No of mol of KCHO2 = 0.2675*0.29 = 0.078 mol
pOH = pkb + log(salt/base)
pkb = -logKa = -log(5.6*10^-4) = 3.252
= 3.252+log(0.078/0.0863)
= 3.21
pH = 14-3.21 = 10.79
after addition of NaOH
= 3.252+log((0.078-0.02)/(0.0863+0.02))
= 2.99
pH = 14-2.99 = 11.01
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