+O e www-awh.aleks.com O ACIDS, BASES AND AQUEOUS EQUILIBRIA Calculating the pH
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+O e www-awh.aleks.com O ACIDS, BASES AND AQUEOUS EQUILIBRIA Calculating the pH at equivalence of a titration Albert A chemist titrates 160.0 m of a 0.7164 M hydrocyanic acid (HCN) solution with 0.7907 M KOH solution at 2s °c. Calculate the pH at equivalence. The pK, of hydrocyanic acid is 9.21. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KoH solution added alo Ar pH = Explanation Check 2017 McGaw Hill Education Al hahts Reserved Terms of Use PrivacyExplanation / Answer
we have below equation to be used:
pKa = -log Ka
9.21 = -log Ka
log Ka = -9.21
Ka = 10^(-9.21)
Ka = 6.166*10^-10
find the volume of KOH used to reach equivalence point
M(HCN)*V(HCN) =M(KOH)*V(KOH)
0.7164 M *160.0 mL = 0.7907M *V(KOH)
V(KOH) = 144.9652 mL
we have:
Molarity of HCN = 0.7164 M
Volume of HCN = 160 mL
Molarity of KOH = 0.7907 M
Volume of KOH = 144.9652 mL
mol of HCN = Molarity of HCN * Volume of HCN
mol of HCN = 0.7164 M * 160 mL = 114.624 mmol
mol of KOH = Molarity of KOH * Volume of KOH
mol of KOH = 0.7907 M * 144.9652 mL = 114.624 mmol
We have:
mol of HCN = 114.624 mmol
mol of KOH = 114.624 mmol
114.624 mmol of both will react to form CN- and H2O
CN- here is strong base
CN- formed = 114.624 mmol
Volume of Solution = 160 + 144.9652 = 304.9652 mL
Kb of CN- = Kw/Ka = 1*10^-14/6.166*10^-10 = 1.622*10^-5
concentration ofCN-,c = 114.624 mmol/304.9652 mL = 0.3759M
CN- dissociates as
CN- + H2O -----> HCN + OH-
0.3759 0 0
0.3759-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 ((1.622*10^-5)*0.3759) = 2.469*10^-3
since c is much greater than x, our assumption is correct
so, x = 2.469*10^-3 M
[OH-] = x = 2.469*10^-3 M
we have below equation to be used:
pOH = -log [OH-]
= -log (2.469*10^-3)
= 2.61
we have below equation to be used:
PH = 14 - pOH
= 14 - 2.61
= 11.39
Answer: 11.39
Feel free to comment below if you have any doubts or if this answer do not work
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