Question

Consider the titration of 40.0 mL of 0.250 M HF with 0.200 M NaOH. How many milliliters of base are required to reach the equivalence point? Calculate the pH at the following points:

a)After the addition of 10.0 mL of base

b)Halfway to the equivalence point

c)At the equivalence point

d)After the addition of 80.0 mL of base

Answer #1

**no of mol of HF = 40/1000*0.25 = 0.01 mol**

**pka of HF = 3.17**

**a)After the addition of 10.0 mL of base**

**NO of mol of NaOH added = 10/1000*0.2 = 0.002
mol**

**pH = pka + log(salt/acid)**

** =
3.17+log(0.002/(0.01-0.002))**

** = 2.57**

**b)Halfway to the equivalence point , the pH of buffer =
pka of acid**

**= 3.17**

**c)At the equivalence point**

**No of mol of HF = NO of mol of NaOH added**

**concentration of salt = 0.01/(0.05+0.04) = 0.11
M**

**pH = 7+1/2(pka + logC)**

** 7+1/2(3.17+log0.11) =
8.1**

**d)After the addition of 80.0 mL of base**

**No of mol of base added = 80/1000*0.2 = 0.016
mol**

**excess No of mol of base added 0.016 - 0.01 = 0.006
mol**

**concentration of excess base = 0.006 / 0.12 = 0.05
M**

**pOH = -log(OH-)**

** = -log0.05 = 1.3**

**pH = 14-1.3 = 12.7**

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