Question

Calculate the pH of each of the following strong acid solutions.

(a) 0.00526 M HCl
pH =  

(b) 0.839 g of HBr in 29.0 L of solution
pH =  

(c) 46.0 mL of 8.80 M HCl diluted to 4.20 L
pH =  

(d) a mixture formed by adding 76.0 mL of 0.00851 M HCl to 23.0 mL of 0.00876 M HBr

pH =

Answer 1

pH is a scale from 1 to 14 to specify aqueous solution as acid or base. It is a negative logarithm of concentration of hydrogen ion in the solution.

pH = -log[H+]

Part a). 0.00526 M HCl

pH= -log[0.00526] = -log[5.26* 10^-3] = -[log5.26+log10^-3]= -[log5.26-3log10]=3log10-log5.26=3*1-0.7209 =2.28

pH=2.28 answer

Part b). 0.839 g of HBr in 29.0 L of solution.

pH is based on concentration of solution, for that we need to know number of moles of solute present in solution

Mole = Mass in gram/molar mass

Moles=0.839 g/80.91 =0.01037 mol

Concentration of HBr= Number of moles of solutes H⁺ present in 1 litre of solution

Concentration =    = moll^-1

pH =

Answer pH=3.4474

Part c) 46.0 mL of 8.80 M HCl diluted to 4.20 L

We need to find conncentration as per dilution equation

M1VI=M2V2

PART d). image