Calculate the pH change that results when 15 mL of 2.6 M HCl is added to 600. mL

ID: 874202 • Letter: C

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 B,C, or D would be greatly appreciated.

(a) pure water - I know that this pH value is -5.8


(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

The pH change is calculated as follows,

(a) 15 mL of 2.6 M HCl is added to 600. mL of pure water

moles of HCl added = M x L

= 2.6 x 0.015

= 0.039 moles

Total volume of solution = volume of HCl solution + volume of pure water

= 0.015 L + 0.600 L

= 0.615 L

Molar concentration of HCl in new solution = moles / L

= 0.039 / 0.615

= 0.063 M

Neglecting the (H^+) from the ionization of water (because it is too small),

pH of new solution = -log (0.063)

= 1.20

(b) 15 mL of 2.6 M HCl is added to 600 mL of 0.10 M CH3COO-1

moles of HCl (as calculated in part (a)) added = 2.6 M x 0.015 L

= 0.039 moles

Initial moles of CH3COO- = M x L

= 0.10 M x 0.600 L

= 0.06 moles

Thus, 0.039 moles of CH3COO- will be neutralized with 0.039 moles of HCl

Remaining moles of CH3COO- in solution = 0.06 - 0.039

= 0.021

New moles of CH3COOH in solution = 0.06 + 0.039

= 0.099 moles

Total volume of new solution = volume of added HCl solution + volume of CH3COO- solution

= 0.015 L + 0.600 L

= 0.615 L

Concentration of CH3COO- in new solution = 0.021 / 0.615

= 0.034 M

Concentration of HA in new solution = 0.099 / 0.615

= 0.16 M

Ka for acetic acid = [CH3COO-] [H+] / [CH3COOH]

Ka for acetic acid = 1.8 x 10^-5

Feeding all the values we get,

1.8 x 10^-5 = (0.034)[H+] / (0.16)

[H+] = 8.5 x 10^-5

pH = -log(8.5 x 10^-5)

= 4.07

Therefore, pH of new solution is 4.07

Change in pH = 4.07 - 4.74

= -0.67

Since CH3COOH is a weak acid, the pH will be determined by the conc. of HCl in solution thus,

pH = 1.20 same as in (a)

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Calculate the pH change that results when 15 mL of 2.6 M HCl is added to 600. mL

Calculate the pH change that results when 15 mL of 2.6 M HCl is added to 600. mL

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