Question

Calculate the pH of the resulting solution if 27.0 mL of 0.270 M HCl(aq) is added to

(a) 37.0 mL of 0.270 M NaOH(aq).

(b) 17.0 mL of 0.370 M NaOH(aq).

Answer 1

Answer:

HCl:

M₁ = 0.27; V₁= 27 mL

a) NaOH :

M2 = 0.27 M ; V₂ = 37 mL

Here HCl and NaOH both are the strong acid and strong Bases

The volume of the NaOH is more and the both concentrations are same. Hence the the resultant solution is Basic in Nature.

Resultant concentration caluculation:

MV = M₂V₂ - M₁V₁

V = Total volume (HCl and NaOH volume)

M (27+37) =(0.27 x 37) - (0.27 x 27)

M x 64 = 9.99 - 7.29

M x 64 = 2.7

M = 2.7 / 64

M = 0.0421

The concentration is the Basic concentration:

P^OH = -log₁₀[-OH]

P^OH = -log₁₀[ 0.0421]

P^OH = -log₁₀[ 421 x 10^-4]

P^OH = 4 - log 421

P^OH = 4 - 2.624

P^OH = 1.367

P^H caluculation:

We know that P^H + P^OH = 14

P^H = 14 - P^OH

P^H = 14 - 1.367

P^H = 12.633

b)

HCl:

M₁ = 0.27; V₁= 27 mL

NaOH :   

M2 = 0.37 V₂ = 17mL

Here HCl and NaOH both are the strong acid and strong Bases

Resultant concentration caluculation:

MV = M1V₁ - M2V2 [ Acid number of moles is more]

V = Total volume (HCl and NaOH volume)

M (27+17)=(0.27 x27)- (0.37 x 17)

M x 44= 7.29 - 6.29

M x44= 1

M = 1/ 44

M = 0.0227

The resultant concentration is Acidic concentration

P^H = -log₁₀[H+]

P^H = -log₁₀[0.0227]

P^H = -log₁₀[227 x 10^-4 ]

P^H = 4 - log 227

P^H = 4 - 2.356

P^H = 1.644