For each of the following solutions, calculate the initial pH and the final pH a

Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.010 mol of NaOH .

A.) For 210.0 mL of pure water, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

B.) For 210.0 mL of a buffer solution that is 0.195 M in HCHO2 and 0.280 M in KCHO2, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

C.)For 210.0 mL of a buffer solution that is 0.295 M in CH3CH2NH2 and 0.275 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

Explanation / Answer

A.)

Pure water : pH = 7.00 (initial)

moles NaOH = 0.010

So,

[OH-]= 0.010 mol/ 0.210 L=0.048 M

pOH = - log 0.048= 1.32

pH = 14 - pOH = 14 - 1.32 = 12.68

B.)

Ka of HCO2H = 1.8 x 10–4

pKa = - log Ka =-log (1.8 x 10–4) = 3.74

using Henderson-hasselbarg equation

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

pH = 3.74 + log 0.280/0.195

      = 3.74 + log (1.44)

      = 3.74 + 0.16

      = 3.90 (INITIAL)

moles formic acid = (0.210 L) x 0.195 M = 0.041 mole

moles formate ion = (0.210 L) x 0.280 M = 0.059 mole

On addition of NaOH the concentarion of HCOOH will decrease and HCOO- will increase.

Moles of NaOH added = 0.010

So after addition of NaOH,

moles formic acid = (0.041 – 0.010) mole = 0.031 mole

moles formate ion = (0.059 + 0.010) mole = 0.069 mole

pH = 3.74 + log (0.069 / 0.031)

     = 3.74 + log (2.23)

     = 3.74 + 0.35

     = 4.09 (final)

C.)

Kb = 5.6 x 10-4

pKb = - log Kb = -log(5.6 x 10-4) = 3.25

using Henderson-hasselbarg equation

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

      = 3.25 + log (0.275/0.295)

      = 3.25 + log (0.93)

      = 3.25 + log (0.93)

      = 3.25 -0.032

      = 3.218 = 3.22

pH = 14 - 3.32 =10.78 ( initial pH)

Now,

moles ethylamine = 0.295 M x 0.210 L= 0.062

moles ethylammonium = 0.275 M x 0.210 = 0.058

When NaOH is added it will react with ethylammonium to give ethylamine.

CH3CH2NH3+ + OH- = CH3CH2NH2 + H2O

Moles of NaOH added = 0.010 mole

So, moles after addition of NaOH will be

moles CH3CH2NH3+ = 0.058 - 0.010 = 0.048

moles CH3CH2NH2   = 0.062 + 0.010 = 0.072

So,

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

      = 3.25 + log (0.048/0.072)

      = 3.25 + log (0.67)

      = 3.25 -0.17

      = 3.08

pH = 14 - pOH = 14 - 3.08 = 10.92 (final)

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