Question

Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 2.67-M HCl to 540. mL of each of the following solutions. Change is defined as final minus initial, so if the pH drops upon mixing the change is negative. a) water pH before mixing = Correct: Your answer is correct. pH after mixing= pH change = b) 0.153 M C2H3O21- pH before mixing = pH after mixing= pH change = c) 0.153 M HC2H3O2 pH before mixing = pH after mixing= pH change = d) a buffer solution that is 0.153 M in each C2H3O21- and HC2H3O2 pH before mixing = pH after mixing= pH change =

Answer 1

Moles of HCL added in each case = 0.010 L * 2.67 M = 0.0267 moles

moles of CH3COO- initially present = 0.54 L * 0.153 M = 0.083 moles

pH of the solution before mixing can be calculated by considering following equilibrium.

CH3COO- + H2O ==> CH3COOH + OH-

as CH3COOH is a weak acid , so its conjugate base is a strong base and will be completely dissociated.

moles of CH3COO- = moles of OH- = 0.153 M

pOH = -log[0.153] = 0.82

pH = 14-0.82 = 13.18

If HCl is added , it will react with HCl to convert it to acetic acid. So, the number of moles of HCl that has been added has to be deducted from moles of CH3COO- initially present.

moles of CH3COO- present after mixing = 0.083-0.0267 = 0.0563 moles

Total volume = 540 mL + 10mL = 550 mL

Molarity = 0.0563 moles*1000mL/550mL = 0.1 M

pOH = -log[OH-] = 1

pH = 14-1 = 13.

change in pH = 13.18-13 = 0.18

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(c) Moles of acetic CH3COOH present = 0.083 moles

CH3COOH ===> CH3COO- + H+

	CH3COOH	CH3COO-	H+
initial	0.153	0	0
change	-x	+x	+x
equilibrium	0.153-x	+x	+x

Ka = [H+][CH3COO-]/[CH3COOH] refer to your text book for ka value of CH3COOH

or, 1.8*10^-5 = x^2/(0.153-x)

or, x =0.0016 M

pH = -log[H+] = 2.79

pH before mixing the acid = 2.79

[H+] present in the before adding any HCl = 0.54 L * 0.0016 M = 8.64*10-4 moles

moles of HCl = moles of H+ added = 0.0267 moles

Total moles of H+ = 0.0267 + 0.000864 = 0.02756 moles

Molarity of H+ after mixing the two solutions = 0.02756 moles*1000mL/550mL =0.050 M

pH after mixing = -log[H+] = -log[0.05] =1.30

change in pH = 2.79-1.30 = 1.49

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(d) ph of a buffer solution can be calculated by using Hinderson hasselbalch equation.

pH = pka + log[CH3COO-/CH3COOH]

or, pH = pKa (as [CH3COO-]=[CH3COOH])

pH = -log ka = -Log (1.8*10^-5) = 4.74

when HCl is added, it will react with CH3COO- to convert it to CH3COOH. So that moles of HCl that has been added has to be deducted from moles of CH3COO- present and has to be added to moles of CH3COOH present.

Moles of CH3COO- pressent after mixing = 0.0563 (previously calculated)

Moles of CH3COOH present after mixing = (0.083-0.0267 = 0.1097

pH = pKa + log [CH3COO-/CH3COOH]

or, pH = 4.74 + log[0.0563/0.1097] = 4.45

pH change = 4.74-4.45 = +0.29