Answer:
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...........(*****)
