PART 1: A buffer solution contains 0.431 M K H 2 PO 4 and 0.289 M K 2 HPO 4 . If

Question

PART 1: A buffer solution contains 0.431 M KH2PO4 and 0.289M K2HPO4.

If 0.0304 moles of sodium hydroxide are added to 125mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume does not change upon adding sodium hydroxide)

pH =

PART 2:A buffer solution contains 0.431 M ammoniumbromide and 0.342 M ammonia.

If 0.0288 moles of potassium hydroxide are added to 125 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume change does not change upon adding potassium hydroxide)

pH =

Explanation / Answer

PART 1:

Moles of KH2PO4 = 0.431 M x 0.125 L = 0.053875 moles
Moles of K2HPO4 = 0.289 M x 0.125 L = 0.036125 moles

Now, you are adding 0.0304 moles of NaOH

This will stoichiometrically react with the KH2PO4 and convert it to K2HPO4. So, after the addition you have:

Moles of KH2PO4 = 0.053875 moles – 0.0304 moles = 0.023475 moles
Moles of K2HPO4 = 0.036125 moles + 0.0304 moles = 0.066525 moles

Total volume = 125 mL = 0.125 L

So, the concentration terms are
[KH2PO4] = 0.023475 moles / 0.125 L = 0.1878 M
[K2HPO4] = 0.066525 moles / 0.125 L = 0.5322 M

Ka for KH2PO4- = 6.2 x 10-8 (From literature, you check with your Ka)

pKa = -log Ka
       = - log (6.2 x 10-8)
       = 7.21

Now,

pH = pKa + log { [salt] / [acid] }

pH = pKa + log { [K2HPO4] / [KH2PO4] }

     = 7.21 + log (0.5322 / 0.1878)

     = 7.21 + log (2.83)

     = 7.55 + 0.45

     = 8.00

PART 2:

Moles of NH3    = 0.342 M x 0.125 L = 0.04275 moles
Moles of NH4Br = 0.431 M x 0.125 L = 0.053875 moles

Now, you are adding 0.0288 moles of KOH

This will stoichiometrically react with the NH4Br and convert it to NH3. So, after the addition you have:

Moles of NH3 = 0.04275 moles + 0.0288 moles = 0.07155 moles
Moles of NH4Br = 0.053875 moles + 0.0288 moles = 0.082675 moles

Total volume = 125 mL = 0.125 L

So, the concentration terms are
[NH3] = 0.07155 moles / 0.125 L = 0.5724 M
[NH4Br] = 0.082675 moles / 0.125 L = 0.6614 M

Kb for NH3 = 1.8 x 10-5 (From literature, you check with your Ka)

pKb = -log Kb
       = - log (1.8 x 10-5)
       = 4.74

Now,

pOH = pKb + log { [salt] / [base] }

pOH = pKb + log { [NH4Br] / [NH3] }

     = 4.74 + log (0.6614 / 0.5724)

     = 4.74 + log (1.16)

     = 4.74 + 0.06

     = 4.80

pH = 14 – pOH
     = 14 – 4.80
      = 9.2

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