Answer to Question #148049 in General Chemistry for shahd massri

Question

calculate the pH of solution after the addition 15 ml of 0.2 M ba(oh)2 to 70ml of 0.1 M hcl

Answer 1

Ba(OH)_2(aq) + 2HCl_(aq) = BaCl_2(aq) + 2H₂O_(l)

Moles of Ba(OH)_2(aq) in 15mL

 =0.2×15/1000

=0.003moles

Moles of HCl in 70mL

=0.1×70/1000

=0.007 moles

Ba(OH)₂: HCl

    1     :   2

Therefore, 0.003moles of Ba(OH)₂ reacted with 2(0.003) moles of HCl

Moles of HCl that reacted= 0.006moles

Excess moles of HCl in the solution= (0.007-0.006) moles= 0.001moles

Molarity of excess HCl in (70+15) mL solution= 1000 × 0.001/85

=0.011765M

[H⁺] =0.011765M

pH=-log[H⁺]

   =-log (0.011765)

   =1.9294