Question

What is the pH of a sulfuric acid solution made by pipetting 25 ml of a concentrated sulfuric acid H2SO4 (98 wt%, F.W 98.0 g/mol) and diluting to 250 ml?

Answer 1

lets calculate the initial concentration of H2SO4
Let volume of solution be 1 L
volume , V = 1 L
= 1*10^3 mL


density, d = 1.0 g/mL
use:
mass = density * volume
= 1 g/mL *1*10^3 mL
= 1000.g
This is mass of solution
mass of H2SO4 = 98.0 % of mass of solution
= 98.0*1000.0/100
= 980.0 g

Molar mass of H2SO4= 98.086 g/mol


mass(H2SO4)= 980 g

use:
number of mol of H2SO4,
n = mass of H2SO4/molar mass of H2SO4
=(980.0 g)/(98.0 g/mol)
= 10.0 mol
volume , V = 1 L


use:
Molarity,
M = number of mol / volume in L
= 10.0/1
= 10.0 M

use dilution formula
M1*V1 = M2*V2
1---> is for stock solution
2---> is for diluted solution

Given:
M1 = 10.0 M
V1 = 25.0 mL
V2 = 250.0 mL

use:
M1*V1 = M2*V2
M2 = (M1*V1)/V2
M2 = (10*25)/250
M2 = 1 M

this is concentration of H2SO4 after dilution

[H+] = 2*[H2SO4] = 2*1 M = 2 M

use:
pH = -log [H+]
= -log (2)
= -0.301
Answer: -0.301