Question

Calculate the pH of each of the following strong acid solutions.
(a) 0.00512 M HBr

pH =  

(b) 0.633 g of HI in 18.0 L of solution

pH =  

(c) 44.0 mL of 3.90 M HBr diluted to 1.30 L

pH =  

(d) a mixture formed by adding 89.0 mL of 0.00215 M HBr to 89.0 mL of 0.000310 M HI

pH =  

Answer 1

pH = -log[H⁺]

(a)  0.00512 M HBr Since it is a strong acid dissociates almost completely. Hence the concentration of H⁺ in mol L^-1 which we generally write as [H⁺] =  0.00512 mol L^-1

pH = -log[H⁺] = -log(0.00512) = -log (5.12x10^-3) = -(-3log10 + log5.12) = 3 - 0.7092 = 2.29

(b) 0.633 g of HI in 18.0 L of solution

In one L 0.633/18 g of HI will be present

That is 0.0351 g of HI in one L

Molar mass of HI is 127.911 g mol^-1

hence, moles of HI per L of the solution is = 0.0351/127.911 = 2.74x10^-4 mol L^-1

pH = -log (2.74x10^-4) = 3.56

(c) 44.0 mL of 3.90 M HBr diluted to 1.30 L

First step is to find out the molarity of the diluted solution

For which we can use the simple formula M₁V₁ = M₂V2 V1 =  44.0 mL = 0.044L M1 = 3.90 M   

and v2 = 1.30L we need to find out M2

3.90 x 0.044 = M2 x 1.30

M2 = 0.132 M

pH = -log[H⁺] = -log 0.132 = 0.8794

(d) a mixture formed by adding 89.0 mL of 0.00215 M HBr to 89.0 mL of 0.000310 M HI

First we need to find out the total H⁺ ion concentration

m mol of HBr = 89 x 0.00215 = 0.1913

Similarly m mol of HI = 89 x0.000310= 0.0275

Total m mol = 0.0275+ 0.1913 = 0.2188

Total moles = 0.2188x10^-3  

These are the total number of moles of HBr and HI in the solution. Because they are strong acids, they dissociate almost completely and hence the concentration of H⁺ ions = 0.2188x10^-3

However these number of moles of H⁺ iona are in 89+89 = 178 ml

Therefore in 1000ml how many H⁺ ions will be there?

= 1000 x0.2188x10^-3 /178 = 1.229x10^-3mol L^-1

Therefore pH = -log (1.229x10^-3) = 2.91