Analytical chemistry, please answer all questions below with FULL SOLUTIONS! Ans

Question

Analytical chemistry, please answer all questions below with FULL SOLUTIONS!

Answers

#7: 6.98

#8: ** not given on problem set, but please show anyways!

#9: 6.98

Question 7 [5 mark| [PROBLEM A2l Using the avity based dio for pH, what is the pH of a 0.05 M NaySO, soion? Compare this pH vale you would get sing the conentration based delinition of pH Question 8 5 markPROBLEM A3] Write out ALL sineous relationships that would deline the equilibrium composition of a weak acid with 1)Ka valile of 3.4 and a total concentration of 1.3G mM. Include activity n your equations. Question 9 [5 mark] IPROBLEI A Using te ativy based definintion for pH, what is the pH of a 0.1 M KCI solution? Compare this pH value you would get using the contration based delinition of pH.

Explanation / Answer

Solution 7.

SO4^2- + H2O <==> HSO4- + OH-

ionic strength of solution (u) = 1/2(Cizi^2)

= 1/2[0.1 x 1^2 + 0.05 x 2^2] = 0.15

acitivty coefficient of [H+]

logY[H+] = -0.51 x 1^2 x sq.rt(0.15)/1 + 900 x sq.rt(0.15)/305)

Y[H+] = 0.81 and Y[OH-] = 0.74

[H+] = inv.sq.rt(1 x 10^-14/0.81 x 0.74) = 1.29 x 10^-7 M

pH = -log(1.29 x 10^-7 x 0.81) = 6.98

Now with the usual method of concentration,

Kb2 = 9.77 x 10^-13 = x^2/0.05

x = [OH-] = 2.21 x 10^-7 M

pOH = -log[OH-] = 6.65

pH = 7.34

Solution 8 : HA <==> H+ + A-

Ka = Y[H+]C[H+].Y[A-]C[A-]/Y[HA].C[HA] = 3.98 x 10^-4

ionic strength = 1/2(0.00136 + 0.00136) = 0.0136

logY[H+] = -0.51 x 1 x sq.rt(0.00136)/1+900 x sq.rt(0.00136)/305)

Y[H+] = 0.96

Y[OH-] = 0.88

3.98 x 10^-4 = x^2(0.96 x 0.88)/0.00136

x = 8 x 10^-4

equilibrium concentration of weak acid = 0.00136 - 8.0 x 10^-4 = 5.60 x 10^-4 M

Solution 9 : 0.1 M KCl

Cl- + H2O ---> HCl + OH-

[H+] = [OH-]

ionic strength = 1/2(0.1 + 0.1) = 0.1

acitvity coefficient Y[H+] = -0.51 x 1 x sq.rt(0.1)/1+900 x sq.rt(0.1)/305)

Y[H+] = 0.83 and Y[OH-] = 0.76

Kw = Y[H+]C[H+].Y[OH-]C[OH-]

[H+]^2 = 1 x 10^-14/(0.83 x 0.76)

[H+] = 1.26 x 10^-7

pH = -log[1.26 x 10^-7 x 0.83) = 6.98

Using concentration formula,

[H+] = 1 x 10^-7 M

pH = 7

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.