Question
Calculate the pH change that results when 15 mL of 2.6 M HCl is added to 600. mL of each of the following solutions. (See the appendix.) Any help on any part of this problem would be greatly appreciated. I can't figure out how to solve B-D at all.
(a) pure water - I KNOW THIS IS -5.8 pH.
(b) 0.10 M CH3COO?
(c) 0.10 M CH3COOH
(d) a solution that is 0.10 M in each CH3COO? and CH3COOH.
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
Solution :-
Given data
15 ml 2.6 M HCl
Part a) 15 ml 2.6 M HCl + 600 ml pure water
After adding HCl solution to 600 ml water the concentration of the HCl will decrease
So lets calculate the concentration of the HCl after mixing
The volume will change to after 615 ml after mixing 15 ml with 600 ml
So using this volume lets calculate new concentration of the HCl
Formula
M1V1=M2V2
Where,
M1 = initial concentration,V1= initial volume, M2= Final concentration, V2= final volume
Lets put the values in the formula and solve for M2
M1V1=M2V2
2.6 M * 15 ml = M2 * 615 ml
M2 = (2.6 M * 15 ml ) / 615 ml
M2 = 0.0634 M
Since HCl is the strong acid so we can directly use the concentration to calculate the pH because strong acid dissociates completely.
Lets calculate pH using this H+ concentration
Formula to calculate pH is as follows
pH= - log[H+]
pH= - log [ 0.0634]
pH= 1.20
part b) 15 ml 2.6 M HCl + 600 ml 0.10 M CH3COO-
reaction equation
HCl + CH3COO-