Q, LON-CAPA Titration ofi × O lor capasa yo kuca/res stuehem215 problems lecture
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Q, LON-CAPA Titration ofi × O lor capasa yo kuca/res stuehem215 problems lecture acidbase/titrwacid sba mL of 0.100-M hydrogen cyanide (Ka-6.20x10-10) is titrated with 0.100-M NaOH What is the initial pH of the hydrogen cyanide solution? Tries 0/5 What is the pH of the solution after 14.0 mL NaOH has been added? Submit Answer Tries o/5 What is the pH of the solution after a total of 17.5 ml NaOH has been added? Submit Answer Tries 0/5 What is the pH of the solution after a total of 21.0 mL NaOH has been added? Submit Answer Tries 0/5 What is the pH of the solution after a total of 35.0 mL NaOH has been added? Submit Answer Tries o/S What is the pH of the solution after a total of 42.0 mL NaOH has been added? Submit Answer Tries 0/s Post Discussion Type here to search scExplanation / Answer
1)when 0.0 mL of NaOH is added
Lets write the dissociation equation of HCN
HCN -----> H+ + CN-
0.1 0 0
0.1-x x x
Ka = [H+][CN-]/[HCN]
Ka = x*x/(c-x)
Assuming small x approximation, that is lets assume that x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((6.2*10^-10)*0.1) = 7.874*10^-6
since c is much greater than x, our assumption is correct
so, x = 7.874*10^-6 M
we have below equation to be used:
pH = -log [H+]
= -log (7.874*10^-6)
= 5.10
2)when 14.0 mL of NaOH is added
we have:
Molarity of HCN = 0.1 M
Volume of HCN = 35 mL
Molarity of NaOH = 0.1 M
Volume of NaOH = 14 mL
mol of HCN = Molarity of HCN * Volume of HCN
mol of HCN = 0.1 M * 35 mL = 3.5 mmol
mol of NaOH = Molarity of NaOH * Volume of NaOH
mol of NaOH = 0.1 M * 14 mL = 1.4 mmol
We have:
mol of HCN = 3.5 mmol
mol of NaOH = 1.4 mmol
1.4 mmol of both will react
excess HCN remaining = 2.1 mmol
Volume of Solution = 35 + 14 = 49 mL
[HCN] = 2.1 mmol/49 mL = 0.0429M
[CN-] = 1.4/49 = 0.0286M
They form acidic buffer
acid is HCN
conjugate base is CN-
Ka = 6.2*10^-10
pKa = - log (Ka)
= - log(6.2*10^-10)
= 9.208
we have below equation to be used:
This is Henderson–Hasselbalch equation
pH = pKa + log {[conjugate base]/[acid]}
= 9.208+ log {2.857*10^-2/4.286*10^-2}
= 9.03
3)when 17.5 mL of NaOH is added
we have:
Molarity of HCN = 0.1 M
Volume of HCN = 35 mL
Molarity of NaOH = 0.1 M
Volume of NaOH = 17.5 mL
mol of HCN = Molarity of HCN * Volume of HCN
mol of HCN = 0.1 M * 35 mL = 3.5 mmol
mol of NaOH = Molarity of NaOH * Volume of NaOH
mol of NaOH = 0.1 M * 17.5 mL = 1.75 mmol
We have:
mol of HCN = 3.5 mmol
mol of NaOH = 1.75 mmol
1.75 mmol of both will react
excess HCN remaining = 1.75 mmol
Volume of Solution = 35 + 17.5 = 52.5 mL
[HCN] = 1.75 mmol/52.5 mL = 0.0333M
[CN-] = 1.75/52.5 = 0.0333M
They form acidic buffer
acid is HCN
conjugate base is CN-
Ka = 6.2*10^-10
pKa = - log (Ka)
= - log(6.2*10^-10)
= 9.208
we have below equation to be used:
This is Henderson–Hasselbalch equation
pH = pKa + log {[conjugate base]/[acid]}
= 9.208+ log {3.333*10^-2/3.333*10^-2}
= 9.21
4)when 21.0 mL of NaOH is added
we have:
Molarity of HCN = 0.1 M
Volume of HCN = 35 mL
Molarity of NaOH = 0.1 M
Volume of NaOH = 21 mL
mol of HCN = Molarity of HCN * Volume of HCN
mol of HCN = 0.1 M * 35 mL = 3.5 mmol
mol of NaOH = Molarity of NaOH * Volume of NaOH
mol of NaOH = 0.1 M * 21 mL = 2.1 mmol
We have:
mol of HCN = 3.5 mmol
mol of NaOH = 2.1 mmol
2.1 mmol of both will react
excess HCN remaining = 1.4 mmol
Volume of Solution = 35 + 21 = 56 mL
[HCN] = 1.4 mmol/56 mL = 0.025M
[CN-] = 2.1/56 = 0.0375M
They form acidic buffer
acid is HCN
conjugate base is CN-
Ka = 6.2*10^-10
pKa = - log (Ka)
= - log(6.2*10^-10)
= 9.208
we have below equation to be used:
This is Henderson–Hasselbalch equation
pH = pKa + log {[conjugate base]/[acid]}
= 9.208+ log {3.75*10^-2/2.5*10^-2}
= 9.38
5)when 35.0 mL of NaOH is added
we have:
Molarity of HCN = 0.1 M
Volume of HCN = 35 mL
Molarity of NaOH = 0.1 M
Volume of NaOH = 35 mL
mol of HCN = Molarity of HCN * Volume of HCN
mol of HCN = 0.1 M * 35 mL = 3.5 mmol
mol of NaOH = Molarity of NaOH * Volume of NaOH
mol of NaOH = 0.1 M * 35 mL = 3.5 mmol
We have:
mol of HCN = 3.5 mmol
mol of NaOH = 3.5 mmol
3.5 mmol of both will react to form CN- and H2O
CN- here is strong base
CN- formed = 3.5 mmol
Volume of Solution = 35 + 35 = 70 mL
Kb of CN- = Kw/Ka = 1*10^-14/6.2*10^-10 = 1.613*10^-5
concentration ofCN-,c = 3.5 mmol/70 mL = 0.05M
CN- dissociates as
CN- + H2O -----> HCN + OH-
0.05 0 0
0.05-x x x
Kb = [HCN][OH-]/[CN-]
Kb = x*x/(c-x)
Assuming small x approximation, that is lets assume that x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.613*10^-5)*5*10^-2) = 8.98*10^-4
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Kb = x*x/(c-x)
1.613*10^-5 = x^2/(5*10^-2-x)
8.065*10^-7 - 1.613*10^-5 *x = x^2
x^2 + 1.613*10^-5 *x-8.065*10^-7 = 0
Let's solve this quadratic equation
Comparing it with general form: (ax^2+bx+c=0)
a = 1
b = 1.613*10^-5
c = -8.065*10^-7
solution of quadratic equation is found by below formula
x = {-b + (b^2-4*a*c)}/2a
x = {-b - (b^2-4*a*c)}/2a
b^2-4*a*c = 3.226*10^-6
putting value of d, solution can be written as:
x = {-1.613*10^-5 + (3.226*10^-6)}/2
x = {-1.613*10^-5 - (3.226*10^-6)}/2
solutions are :
x = 8.9*10^-4 and x = -9.061*10^-4
since x can't be negative, the possible value of x is
x = 8.9*10^-4
[OH-] = x = 8.9*10^-4 M
we have below equation to be used:
pOH = -log [OH-]
= -log (8.9*10^-4)
= 3.0506
we have below equation to be used:
PH = 14 - pOH
= 14 - 3.0506
= 10.95
6)when 42.0 mL of NaOH is added
we have:
Molarity of HCN = 0.1 M
Volume of HCN = 35 mL
Molarity of NaOH = 0.1 M
Volume of NaOH = 42 mL
mol of HCN = Molarity of HCN * Volume of HCN
mol of HCN = 0.1 M * 35 mL = 3.5 mmol
mol of NaOH = Molarity of NaOH * Volume of NaOH
mol of NaOH = 0.1 M * 42 mL = 4.2 mmol
We have:
mol of HCN = 3.5 mmol
mol of NaOH = 4.2 mmol
3.5 mmol of both will react
excess NaOH remaining = 0.7 mmol
Volume of Solution = 35 + 42 = 77 mL
[OH-] = 0.7 mmol/77 mL = 0.0091 M
we have below equation to be used:
pOH = -log [OH-]
= -log (9.091*10^-3)
= 2.0414
we have below equation to be used:
PH = 14 - pOH
= 14 - 2.0414
= 11.96
Feel free to comment below if you have any doubts or if this answer do not work. I will edit it if you let me know
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