Question

What is the [OH-] in th final solution prepared by mixing 20.0 ml of 0.050 M HCl with 30.0 ml of 0.10M Ba(OH)2 ?​

Answer 1

Answer:

Millimoles of H+ produced = 20 x 0.05 = 1

Millimoles of OH- produced = 30 x 0.1 x 2 = 6 (... Each Ba(OH)2 gives 2OH-.)

. .. 

Millimoles of OH- remaining in solution  = 6 - 1 = 5

Total volume of solution = 20 + 30 = 50 mL . .. [OH-] = 5/50 = 0.1 M

Answer 2

Answer:

2HCl+Ba(OH)2-->BaCl2+2H2O

V Of HCl=20ml

0.05M

moles= 1milli mole(as volume in ml)

now moles of Ba(OH)2 = 3 milli mole

now we will find out limiting reagent i.e. HCl

because its moles/ stochiometric ratio is small

so it is a L.R.

Now..

2 mole HCl--> 1mole Ba(OH)2

1 mole HCl--> 1/2 mole Ba(OH)2

remaining moles of Ba(OH)2 are 3(given) - 1/2 (jo react ho chuke hai)

= 2.5 milli mole ( jo bch jayenge)

now we have to find out the remaining moles means jo reaction k baad solution me rh jayenge...

then..

Ba(OH)2---> Ba2+ + 2OH-

now Ba(OH)2 has 2.5 moles remaining

2.5moles Ba(OH)2--> 2mole OH-

1 mole Ba(OH)2-->2x2.5

=5milli. mole

now u can find OH- molarity by..

5milli mole/50ml

=0.10M

.....