What is the pH of 10-9 M HCl?
Here , [H+] for acid is < 10 -7 { which is not possible }
so,
total [H+] = 10 -7 + 10 -9
= 1.01 x 10 -7
Thus, pH = - log [H+]
= 6.99
A commercial sample of sulphuric acid has a specific gravity of 1.8. 10 mL of this acid was diluted up to 1 liter with water. 10 mL of diluted acid required 30 mL of N/10 NaOH for complete neutralization. Calculate the percentage purity of H2SO4 in the commercial sample.
given that , 30 ml of N/10 NaOH was required ,
Applying Normality equation,
N1V1 = N2V2
N1V1 = ( 30 / 1000 ) x (1/10) ------1
Again, 10 ml of diluted acid acid was taken
N1 = 30 / 10 x (1/10)
N1 = 0.3
This means the normality of diluted acid was 0.3N
Since 10 ml of the solution was diluted ,
Mass of acid in 10 ml ,
mass = NEVml /1000
=0.3 x ( 98/2 ) x 10 / (1000)
= 0.147g
Since the total solution was of 1 litre,
Total mass of acid = ( 0.147 / 10 ) x 1000
= 14.7 g
Percentage purity (%C) = (real mass / assumed mass) x 100 %
= ( 14.7 / density x volume )x 100 %
= ( 14.7 / 1.8 x 100 ) x 100 %
= 81. 66%