1. What is the pH of the solution obtained by mixing 50.00 mL of 0.1250 M HOAc a

Question

1. What is the pH of the solution obtained by mixing 50.00 mL of 0.1250 M HOAc and 25.00 mL of 0.1000 M NaOH?

2. Calculate the pH of a solution prepared by mixing 10.0 mL of 0.500 M NaOH and 20.0 mL of 0.50 M benzoic acid solution. (Benzoic acid is monoprotic; its ionization constant is 6.7 x 10^-5).

3. If Ka is 1. 85 x 10^-5 for acetic acid, calculate the pH at one-half the equivalence point and at the equivalence point for a titration of 50 mL of 0.100M acetic acid with 0.100 M NaOH. ( Hint hydrolysis is possible).

Explanation / Answer

1) moles of HoAc = molarity x volume /1000

moles of H0Ac = 0.125 x 50 /1000 = 6.25 x 10-3

moles of NaOH = 0.1 x 25 /1000 = 2.5 x 10-3

the reaction is

NaOH + HOAc ---> NaOAc + H20

moles of HOAc reacted = moles of NaOH = 2.5 x 10-3

moles of NaOAc formed = moles of HOAc reacted = 2.5 x 10-3

new moles of HOAc = 6.25 x 10-3 - 2.5 x 10-3 = 3.75 x 10-3

so now HOAc and NaOAc forms a acidic buffer

according to hasselbach equation

pH = pka + log [salt / acid ]

pH = pKa + log [ NaOAc / HOAc]

pH = 4.76 + log [ 2.5 x 10-3 / 3.75 x 10-3 ]

pH = 4.584

so the pH is 4.584

2)

moles of Benzoic acid = molarity x volume /1000

moles of benzoic acid = 0.5x 20 /1000 = 10 x 10-3

moles of NaOH = 0.5 x 10 /1000 = 5 x 10-3

the reaction is

NaOH + benzic acid ---> sodium benzoate + H20

moles of benzoic acid reacted = moles of NaOH = 5x 10-3

moles of sodium benzoate formed = moles of benzoic acid reacted = 5 x 10-3

new moles of benzoic acid = 10 x 10-3 - 5 x 10-3 = 5 x 10-3

so now benzoic acid and sodium benzoate forms a acidic buffer

according to hasselbach equation

pH = pka + log [salt / acid ]

pH = pKa + log [ sodium benzoate / benzoic acid]

pH = 4.202 + log [ 5 x 10-3 / 5 x 10-3 ]

pH = 4.202

so the pH is 4.202

3)
at equilvalce point

Ma Va = Mb Vb

0.1 x 50 = 0.1 x Vb

Vb = 50 ml

pKa = -log Ka

pKa = -log 1.85 x 10-5

pKa = 4.7328

pH at half neutralization point = pKa

so pH = 4.7328 at half equivalnce point

at equivalnce point

both acid and base completely react to form salt

NaOH + CH3COOH ----> CH3COONa + H20

moles of CH2COONa formed = moles of CH3COOH reacted

moles of CH3COONa formed = 50 x 0.1 / 1000 = 5 x 10-3

total volume = 50 + 50 = 100 ml

[CH3COONa ] = moles x 1000 / volume

[CH3COONa] = 5 x 10-3 x 1000 /100

[CH3COONa] = 0.05

now CH3COONa undergoes hydrolysis

CH3COO- + H20 ----> CH3COOH + OH-

now

[OH-] = sqrt ( Kb x C )

[OH-] = sqrt ( Kw x C / Ka ]

[OH-] = sqrt ( 10-14 x 0.05 / 1.85 x 10-5 ]

[OH-] = 5.198 x 10-6

pOH = -log [ OH-]

pOH = -log 5.198 x 10-6

pOH = 5.284

pH = 14 - pOH

pH = 14 - 5.284

pH = 8.716

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