Question

A 30.0-mL sample of 0.165 M propanoic acid is titrated with 0.300 MKOH.

A)Calculate the pH at 0 mL of added base. D)Calculate the pH at the equivalence point. G)Calculate the pH at 25 mL of added base.

B)Calculate the pH at 5 mL of added base. E)Calculate the pH at one-half of the equivalence point.

C)Calculate the pH at 10 mL of added base. F)Calculate the pH at 20 mL of added base.

Answer 1

The reaction can be expressed as follows (Pr = Propyl group):

PrOOH + KOH = PrOOK + H₂O

The information we have:

• V PrOOH = 30.0mL = 0.03L
• M PrOOH = 0.165 M
• M KOH = 0.300M
• pKa PrOOH = 4.87 (from tables).

Before we start, we're going to need the number of moles of PrOOH present in the sample, for this, use the formula:

M = n/V

n = MV

n = (0.165M)(0.03L)

n = 0.00495 moles PrOOH

A) Calculate the pH at 0 mL of added base.

For this we can use the formula:

pH = -log(Ka*C)^1/2

Where Ka = 10^pKa and C = initial concentration of the acid (we can use this formula becuse the ratio Ka/C < 0.011, if not, we would need to use the quadratic form).

pH = -log(10^-4.87*0.165)^1/2

pH = 2.826

B) Calculate the pH at 5 mL of added base (total volume of solution = 35mL = 0.035L).

First, calculate moles of KOH in 5mL (0.005L) using the same formula as above:

n = MV

n = (0.300M)(0.005L)

n = 0.0015 moles KOH

Sketch an ICE table:

	PrOOH	+ KOH	= PrOOK	+ H2O
I	0.00495	0.0015
C	0.0015	0.0015	0.0015	0.0015
E	0.00345	E	0.0015	0.0015

We can see we have a buffer (weak acid/salt) so we can use the formula:

pH = pKa + log (salt/acid)

pH = 4.87 + log (0.0015 / 0.00345)

pH = 4.51

So, the pH after adding 5mL of base is 4.51

C) Calculate the pH at 10 mL of added base (total volume of solution = 40mL = 0.04L).

First, calculate moles of KOH in 10mL (0.01L) using the same formula as above:

n = MV

n = (0.300M)(0.01L)

n = 0.003 moles KOH

Sketch the ICE table:

	PrOOH	+ KOH	= PrOOK	+ H2O
I	0.00495	0.003
C	0.003	0.003	0.003	0.003
E	0.00195	E	0.003	0.003

We can see we have a buffer (weak acid/salt) so we can use the formula:

pH = pKa + log (salt/acid)

pH = 4.87 + log (0.003 / 0.00195)

pH = 5.06

So, the pH of the solution after adding 10mL of base is 5.06

D) Calculate the pH at the equivalence point.

At the equivalence point nPrOOH = nKOH, so nKOH = 0.00495 moles, but we need to find the volume of KOH that contains those moles first, use the formula:

n = MV

V = n/M

V = 0.00495moles / 0.300M

V = 0.0165 L = 16.5mL

So, the total volume of the solution would be 16.5mL + 30mL = 46.5mL

Sketch the ICE table:

	PrOOH	+ KOH	= PrOOK	+ H2O
I	0.00495	0.00495
C	0.00495	0.00495	0.00495	0.00495
E	E	E	0.00495	0.00495

We see we have a salt (in this case it is a basic salt because it comes from the combination of a weak acid and a strong base), so we can calculate its pH using the formula of pH for a weak base:

pH = 14 + log(Kw/Ka*Csalt)^1/2

pH = 14 + log((10^-14/10^-4.87)*(0.00495mol/0.00465L))^1/2

pH = 9.45

So, the pH of the solution at the equivalence point is 9.45