Question
Calculate the pH change that results when 12 mL of 5.9 M NaOH is added to 753 mL of each the following solutions. (See the appendix.) Any help on B,C, or D would be greatly appreciated! I am lost on these..
(a) pure water - I know that this pH is 5.97 (this answer was marked correct)
(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
Molarity of NaOH = 5.9 M
Volume of NaOH = 12 mL
Moles of NaOH = 5.9*12/1000 = 0.0708 moles
(a). Volume of pure water added = 753 mL
Volume of solution = 765 mL
[OH-] = 0.0708/ 0.765
= 0.0925
pOH = - log[OH-]
= - log 0.0925
= 1.03
pH = 14 - 1.03
= 12.97
(b). 753 mL of 0.10 M NH4Cl
pKa for NH4+ = 9.25
pH = pKa + log [base] /[salt]
= 9.25 + log(5.9/0.10)
= 9.25 + 1.77
= 11.02