Question

What is the [OH-]in the final solution prerared by mixing 20.0 ml of 0.050 M HCL with 30.0mL of 0.10 M Ba (OH)

Answer 1

The balanced chemical equation for reaction of Ba(OH)2 and HCl is as follows:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
​1 mole Ba(OH)2 produce 2 moles OH- ions.
Number of moles of OH- ion in solution = concentration of Ba(OH)2 x volume of Ba(OH)2 x 2
= 0.1 mol L-1 x 0.03 L x 2
= 0.006 mol
Number of moles of H+ ion in solution = Concentration of HCl x volume of HCl
​ = 0.05 mol L-1 x 0.02 L
​ = 0.001 mol

Now after mixing 0.001 mol H+ will reacts with 0.001 mol OH-.
Number of moles of OH- left unreacted = 0.006 - 0.001
= 0.005 mol
Total volume of solution = 20 ml + 30 ml = 50 ml = 0.05 L
​Concentration of OH- ion in final solution = 0.005 mol / 0.05 L = 0.1 mol L-1.

Answer 2

Answer:  0.10 M

Explanation: we know that no. of OH- in a mixture of acid and base

= NV of base -NV of acid ÷ total volume   -------( if resultant solution is                

                                                                                        basic)

from question,

normality of base = 0.1 ×2 = 0.2

so , total OH- =  30×0.2 - 20×0.05 ÷ 50

                      = 5 ÷ 50

                     = 0.1