Consider the titration of 50.0 mL of 0.10 M Holoa, with 0.15 M KoH. What is the

Question

Consider the titration of 50.0 mL of 0.10 M Holoa, with 0.15 M KoH. What is the pH after the addition of 20.0 mL of KOH? H A H A H20 a) 1.30 b) 1.54 2. S c) 3.33 d) 7.00 e) 12.70 A 50.0 mL sample of 0.200 M HBr solution is t with 0.200 M NaoH. What is the pH of he solution after adding 40.0 mL of base? a) 1.40 b) 1.30 c) 3.27 d) 1.65 e) 4.36 Consider the titration of 25.00 mL 0.150 M acetic acid (CH3COOH, Ka 1.8 x 10-5) with 0.100 M NaOH. What is the pH of after the addition of 10.00 mL of NaOH? (2 s mu) Lo a) 5.05 b) 5.18 c) 4.31 d) 4.44 e) 2.92 of 300.0 mL of A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. What is the pH of the solution after the addition KOH? (Ka of HF is 3.5 x 104.) a) 12.40 b) 9.33 c). 5.06 d) 8.94 e) 12.00

Explanation / Answer

1)

we know that

moles = concentration x volume(ml) / 1000

so

moles of HCl04 taken = 0.1 x 50 / 1000 = 5 x 10-3

moles of KOH added = 0.15 x 20 / 1000 = 3 x 10-3

now

the reaction is

HCl04 + KOH --->KCl04 + H20

we can see that

moles of HCl04 reacted = moles of KOH added = 3 x 10-3

moles of KCl04 formed = moles of KOH added = 3 x 10-3

now

finally

moles of HCl04 remaining = 5 x 10-3 - 3 x 10-3 = 2 x 10-3

moles of KOH remaining = 3 x 10-3 - 3 x 10-3 = 0

moles of KCl04 = 3 x 10-3

now

KCl04 is a nuetral salt because it is formed from strong acid HCl04 and strong base KOH

so

pH is contributed by HCl04

now

final volume = 50 + 20 = 70 ml

concentration = moles x 1000 / volume (ml)

[HCl04] = 2 x 10-3 x 1000 / 70

[HCl04] = 0.02857

now

HCl04 ---> H+ + Cl04-

as HCl04 is a strong acid , 100 % dissociation

[H+] = 0.02857

now

pH = -log [H+]

pH = -log 0.02857

pH = 1.544

so

the answer is B) 2.54

2)

we know that

moles = concentration x volume(ml) / 1000

so

moles of HBr taken = 0.2 x 50 / 1000 = 0.01

moles of NaOH added = 0.2 x 40 / 1000 = 0.008

now

the reaction is

HBr + NaOH ---> NaBr + H20

we can see that

moles of HBr reacted = moles of NaOH added = 0.008

moles of NaBr formed = moles of NaOH added = 0.008

now

finally

moles of HBr remaining = 0.01 - 0.008 = 2 x 10-3

moles of NaOH remaining = 8 x 10-3 - 8 x 10-3 = 0

moles of NaBr = 8 x 10-3

now

NaBr is a nuetral salt because it is formed from strong acid HBr and strong base NaOH

so

pH is contributed by HBr

now

final volume = 50 + 40 = 90 ml

concentration = moles x 1000 / volume (ml)

[HBr] = 2 x 10-3 x 1000 / 90

[HCl04] = 0.022222

now

HBr ---> H+ + Br-

as HBr is a strong acid , 100 % dissociation

[H+] = 0.022222

now

pH = -log [H+]

pH = -log 0.022222

pH = 1.65

so

the answer is D) 1.65

3)

we know that

moles = concentration x volume(ml) / 1000

so

moles of CH3COOH taken = 0.15 x 25 / 1000 = 3.75 x 10-3

moles of NaOH added = 0.1 x 10 / 1000 = 1 x 10-3

now

the reaction is

CH3COOH + NaOH ---> CH3COONa + H20

we can see that

moles of CH3COOH reacted = moles of NaOH added = 1 x 10-3

moles of CH3COONa formed = moles of NaOH added = 1 x 10-3

now

finally

moles of CH3COOH remaining = 3.75 x 10-3 - 1 x 10-3 = 2.75 x 10-3

moles of NaOH remaining = 1 x 10-3 - 1 x 10-3 = 0

moles of CH3COONa = 1 x 10-3

now

CH3COOH is a weak acid and CH3COONa is salt of its conjugate base

so

they form a buffer solution

we know that

for buffers

pH = pKa + log [conjugate base / acid]

also

pKa = -log Ka

so

pH = -log Ka + log [CH3COONa / CH3COOH]

now

concentration = moles / volume (L)

as the final volume is same for both CH3COONa and CH3COOH

we get

ratio of concentrations = ratio of moles

so

pH = -log 1.8 x 10-5 + log [ 1 x 10-3 / 2.75 x 10-3]

pH = 4.3054

so

the answer is C) 4.31

4)

we know that

moles = concentration x volume(ml) / 1000

so

moles of HF taken = 0.2 x 100 / 1000 = 0.02

moles of KOH added = 0.1 x 300 / 1000 = 0.03

now

the reaction is

HF + KOH --->KF + H20

we can see that , HF is the limiting reagent

so

moles of KOH reacted = moles of HF added = 0.02

moles of KF formed = moles of HF added = 0.02

now

finally

moles of HF remaining = 0.02 - 0.02 = 0

moles of KOH remaining = 0.03 - 0.02 = 0.01

moles of KF = 0.02

now

KF is a basic salt because it is formed from weak acid HF and strong base KOH

so

pH is contributed by both KF and KOH

now

final volume = 300 + 100 = 400 ml

concentration = moles x 1000 / volume (ml)

[KOH] = 0.01 x 1000 / 400

[KOH] = 0.025

then

[KF] = 0.02 x 1000 / 400

[KF] = 0.05

now

KOH ---> K+ + OH-

as KOH is a strong base , 100 % dissociation

[OH-] = 0.05

now

for KF , it is a weak base

F- + H20---> HF + OH-

Kb = [HF] [OH-] / [F-]

we know that

Kb = 10-14 / Ka = 10-14 / ( 3.5 x 10-4) = 2.857 x 10-11

now

F- + H20 --> HF + OH-

using ICE table

at equilibrium

[HF] = [OH-] = y

[F-] = 0.05 - y

so

Kb = [HF] [OH-] / [F-]

2.857 x 10-11 = [y] [y] / [0.05-y]

as Kb value is very low , y value will be very low so 0.05-y ----> 0.05

2.857 x 10-11 x 0.05 = y2

y = 1.19523 x 10-6

so

[OH-] = 1.19523 x 10-6 M

now

total [OH-] = [OH-] from KOH + [OH-] from KF

so

total [OH-] = 0.025 + (1.19523 x 10-6)

total [OH-] = 0.025001195

now

p0H = -log [OH-]

pOH = -log 0.025001195

p0H = 1.6

now

pH = 14 - pOH

pH = 14 - 1.6

pH = 12.4

so

the answer is A) 12.4

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