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=1.8⋅10−4).

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=5.6⋅10−4).

Express your answers using two decimal places separated by a comma.

I CAN'T FIGURE THIS ONE OUT!

Answer 1

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