Consider the titration of 30.0 mL of 0.300 M HClO4 with 0.150 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. 0.0 mL pH is 15.0 mL pH is 30.0 mL pH is 60.0 mL pH is 80.0 mL pH is.

When Vb = 0.0, pH =0.52

When Vb = 15mL, pH =0.82

When Vb = 30mL, pH = 1-12

When Vb = 60mL, pH = 7

When Vb = 80mL, pH = 12.73

Explanation:

Acid (a) = HClO4

Base (b) = KOH

Given Va = 30ml = 0.03L,

Ca = 0.300M

Cb = 0.150

We need to produce PH for the volumes Vb = 0.0ml, 15ml, 30ml, 60ml, 80ml.

The reaction equation is :

HClO4 + KOH > KClO4 + H2O

From the reaction, the reacting ratio is 1 :1, meaning the reagents requires equal mole proportionately, the reagent in excess determine the PH

Let n represent mole

n(HClO4) = conc. × vol

= 0.3 × 0.03

= 0.009 mol

So when the vol(b) =0.0 only HClO4 is in solution and Can = 0.3M

pH = -log[H]

pH = - log[0.3]

pH = 0.5229 ~= 0.52(*)

When V(b) = 15mL = 0.15

n(KOH) = 0.15 × 0.015

= 0.00225 mol

Hence HClO4 is in excess by =

0.009 - 0.00225 = 0.00675mol

Total vol. solution= 15+30 = 45mL

[H+] = (0.00675/0.045) = 0.15

pH = -log[0.15]

pH ~= 0.82(**)

When Vb = 30mL

n(KOH) = conc. × vol

= 0.150 × 0.03

= 0.0045 mole

Excess n(HClO4) = 0.009 - 0.045

= 0.0045 mole

Total vol = 30 +30 =60mL= 0.060L

[H+] = 0.0045/0.060 =0.075

pH = -log[0.075]

pH ~=1.12(***)

When Vb = 60 ml =0.60L

n(KOH) = 0.15 × 0.06 = 0.009mol

Now thee are equal mole of HClO4 and KOH in solution, the solution is assume neutral or approximately neutral

pH=7()

When Vb= 80mL = 0.080L

n(KOH) = 0.15 ×0.08 = 0.012

The KOH is in excess by = 0.012 - 0.009= 0.003mol

Vol. of solution = 110mL = 0.110L

[OH-] = 0.003/0.11 = 0.027

pOH = -log[OH] = 1.57

pH = 14 - 1.57 =12.73()

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