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
(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. 12 mL of 5.9 M NaOH is added to 753 mL water
so we have diluted the solution
M1V1 = M2V2
5.9 x 12 = 753 X M2
M2 = 0.0940 M
so [OH-] 0.0940
so pOH = 1.02
so pH = 12.97
(Your answer is wrong as pH of base can never be less than 7)
b. On additon of NH4Cl
It will react with NaOH to form NH4OH
NaOH + NH4Cl --> NH4OH + NaCl
mole of NaOH = 12 x 5.9 / 1000 = 0.0708
Moles of NH4Cl = 753 X 0.1 / 1000 = 0.0753
so moles of NH4OH produced will be= 0.0708
moles of NH4Cl left = 0.0753 - 0.0708 = 0.0045 moles
now they will form a bufeer whose pOH can be calculated as
pOH = pKb + log [salt / base] = 4.75 = log [0.0045 / 0.0708 ] = 3.55
pH = 10.44
c. On addition of 753mL of NH3 there will be hardly any change in pH as ammonia is a weak base so the pH will be due to strong base only.
So pH will be = 12.97