pH uni Consider the titration of 20.0 mL of 0.117-M of KX with 0.160-M HCl. The
ID: 553613 • Letter: P
Question
pH uni Consider the titration of 20.0 mL of 0.117-M of KX with 0.160-M HCl. The pKa of HX = 6.00. Give all pH values to 0.01 pH units. a) What is the pH of the original solution before addition of any acid? pll b) How many m of acid are required to reach the equivalence point? Va= mL e) What is the pH at the equivalence point? pll d) What is the pH of the solution after the addition of 8.8 mL of acid? pH = i1 e) What is the pH of the solution after the addition of 17.S mL of acid? til pH = Submit Answer Sawe Progres -19 points A 0.970-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 65.0 mL of this solution was titrated with 0.06577-M NaOH. The pH after the addition of 19.62 mL of base was 3.56, and the equivalence point was reached with the addition of 40.96 mL of base. a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution mmol acid b) What is the molar mass of the acid? g/mol c) What is the pk, of the acid?Explanation / Answer
pH of the solution before any acid added,
pH = 7 + 1/2(pKa + logC)
pH = 7 + 1/2(6 + log(20*0.117/1000)
pH = 8.68
After adding HCl solution and at equivalence point,
number of moles of KX = number of moles of HX
20 * 0.117 = V*0.16
V = 14.625 mL
at equivalence point [KX] = [HCl]
then pH = pKa
pH = 6.00
number of moles of HCl = 8.8 * 0.16/1000 = 1.408 * 10^-3
number of moles of KX = 20 * 0.117/1000 = 2.34 * 10^-3
number of moles of KX remaining = 2.34 * 10^-3 - 1.408 * 10^-3
= 9.22 * 10^-4
pH = 7 + 1/2(pKa + logC)
pH = 7 + 1/2(6 + log(9.22*10^-4)
pH = 8.48
number of moles of HCl = 17.5 * 0.16/1000 = 2.8 * 10^-3
number of moles of KX = 20 * 0.117/1000 = 2.34 * 10^-3
number of moles of HCl remaining = 2.8 * 10^-3 - 2.34 * 10^-3
= 4.6 * 10^-4
pH = -logC
pH = -log(4.6*10^-4)
pH = 3.34
number of moles of base added to equivalence point = 0.06577*40.96/1000 = 2.694*10^-3 moles
1 moles of Base utilizes 1 mole of acid
number of moles of acid in 65 mL of solution = 2.694 * 10^-3 moles
number of moles in original solution = 2.694 * 10^-3*100/65 = 4.145 * 10^-3 moles
molar mass of acid = 0.97/(2.694*10^-3) = 360.05 g/mol
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