Calculate the pH change that results when 12 mL of 5.9 M NaOH is added to 763 mL

Question

Calculate the pH change that results when 12 mL of 5.9 M NaOH is added to 763 mL of each the following solutions. (See the appendix.)

(a) pure water
  

(b) 0.10 M NH4Cl


(c) 0.10 M NH3


(d) a solution that is 0.10 M in each NH4+ and NH3

Acid Name ConjugateAcid Ka pKa Conjugate Base Base Name perchloric acid HClO4 >>1 < 0 ClO41- perchlorate ion hydrohalic acid HX (X=I,Br,Cl) >>1 < 0 X1- halide ion sulfuric acid H2SO4 >>1 < 0 HSO41- hydrogen sulfate ion nitric acid HNO3 >>1 < 0 NO31- nitrate ion hydronium ion H3O1+ 1.0 0.00 H2O water iodic acid HIO3 0.17 0.77 IO31- iodate ion oxalic acid H2C2O4 5.9 x 10-2 1.23 HC2O41- hydrogen oxalate ion sulfurous acid H2SO3 1.5 x 10-2 1.82 HSO31- hydrogen sulfite ion hydrogen sulfate ion HSO41- 1.2 x 10-2 1.92 SO42- sulfate ion phosphoric acid H3PO4 7.5 x 10-3 2.12 H2PO41- dihydrogen phosphate ion hydrofluoric acid HF 7.2 x 10-4 3.14 F1- fluoride ion nitrous acid HNO2 4.0 x 10-4 3.40 NO21- nitrite ion lactic acid HC3H5O3 6.4 x 10-5 3.85 C3H5O31- lactate ion formic acid HCHO2 1.8 x 10-4 3.74 CHO21- formate ion hydrogen oxalate ion HC2O41- 6.4 x 10-5 4.19 C2O42- oxalate ion hydrazoic acid HN3 1.9 x 10-5 4.72 N31- azide ion acetic acid HC2H3O2 1.8 x 10-5 4.74 C2H3O21- acetate ion carbonic acid H2CO3 4.3 x 10-7 6.37 HCO31- hydrogen carbonate ion hydrogen sulfite ion HSO31- 1.0 x 10-7 7.00 SO32- sulfite ion hydrosulfuric acid H2S 1.0 x 10-7 7.00 HS1- hydrogen sulfide ion dihydrogen phosphate ion H2PO41- 6.2 x 10-8 7.21 HPO42- hydrogen phosphate ion hypochlorous acid HClO 3.5 x 10-8 7.46 ClO1- hypochlorite ion ammonium ion NH41+ 5.6 x 10-10 9.25 NH3 ammonia hydrocyanic acid HCN 4.0 x 10-10 9.40 CN1- cyanide ion hydrogen carbonate ion HCO31- 4.7 x 10-11 10.33 CO32- carbonate ion hydrogen phosphate ion HPO42- 4.8 x 10-13 12.32 PO43- phosphate ion hydrogen sulfide ion HS1- 1.3 x 10-13 12.89 S2- sulfide ion water H2O 1.0 x 10-14 14.00 OH1- hydroxide ion ammonia NH3 <<10-14 NH21- amide ion hydroxide ion OH1- <<10-14 O2- oxide ion

Explanation / Answer

(a) pH of pure water = 7

[NaOH] after addition to water = 5.9 M * 12mL /( 12+ 763) mL

= 0.091 M

pOH = -log[OH-]

= - log 0.091

= 1.04

pH = 14- pOH = 12.96

pH change = dPH = 12.96-7 = 5.96

(b) NH4Cl is a salt of NH3 (pKb = 4.75) and HCl.

It will hydrolyze in the solution, the resulting solution is acidic.

pOH= 0.5( pKw + pKb + logC)

= 0.5 (14+4.75 + log 0.1)

= 7.375

pH = 6.625

On addition of NaOH, it will react with NH4Cl:

NH4Cl + NaOH = NH4OH + NaCl

Total volume of the solution = 12 + 763 = 775 mL

Concentration of NaOH in 775 mL solution (before reaction) = 12 mL * 5.9 M / 775 mL = 0.091 M

Concentration of NH4Cl in 800 mL solution (before reaction) = 763 ml * 0.1 M/ 775 mL = 0.098 M

NaOH will react completely with NH4Cl.

After reaction, [NH4OH] = 0.091 M

After reaction, [NH4Cl] = 0.098-0.091 = 0.007 M

The solution will act as a buffer.

pOH = pKb + log [NH4Cl]/[NH4OH] = 4.75 + log 0.007/0.091

= 3.64

pH = 10.36

dpH = 10.36 - 6.625 =3.73

(c) pOH of 0.1 M NH3 = 0.5 (pKb - logC)

= 0.5(4.75 - log0.1)

= 2.875

pH = 11.125

NH3 gives OH- in solution: NH3 + H2O = NH4+ + OH-

Kb = 1.77 * 10-5 .

[NaOH] after addition = 0.091 M (see previous calculation)

When NaOH is added in the solution, it dissociate completely to give OH- in the solution.

As it is much stronger base than ammonia, [OH-] from ammonia can be neglected in the solution.

[OH-] in the solution 0.091 M

pH = 12.96

dPH = 12.96-11.125

= 1.835

(d) this is a buffer solution,

pOH = pKb + log [NH4+]/[NH3]

= 4.75 + log 0.1/0.1

= 4.75

pH = 9.25

OH- from NaOH reacts with NH4+ to give NH3, so concentration of NH4+ decreases and concentration of NH3 increases.

OH- + NH4+ = NH3 + H2O

Concentration of NaOH in 775 mL solution (before reaction) = 12 mL * 5.9 M / 775 mL = 0.091 M

Concentration of NH4+ in 800 mL solution (before reaction) = 763 ml * 0.1 M/ 775 mL = 0.098 M

Concentration of NH3 in 800 mL solution (before reaction) = 763 ml * 0.1 M/ 775 mL = 0.098 M

After reaction, [NH4+] = 0.098 M - 0.091M = 0.007 M

After reaction, [NH3] = 0.098 M + 0.091M = 0.189 M

pOH = pKb + log [NH4+]/[NH3]

= 4.75 + log 0.007/0.189

= 3.32

pH = 14-3.32 = 10.68

dpH = 10.68 - 9.25 = 1.43

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