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In: Chemistry

How many mL of 0.125 M Ba(OH)2 would be required to completely neutralize 75.0 mL of...

How many mL of 0.125 M Ba(OH)2 would be required to completely neutralize 75.0 mL of 0.845 M HCl? What is the pH of the solution at the equivalence point?

Expert Solution

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

1 mole 2 moles

Ba(OH)2 HCl

M1 = 0.125M M2 = 0.845M

V1 = V2 = 75ml

n1 = 1 n2 = 2

M1V1/n1 = M2V2/n2

V1 = M2V2n1/M1n2

= 0.845*75*1/0.125*2 = 253.5ml

normality of Ba(OH)2 = n* molarity

= 2*0.125 =0.25N

normality of HCl = n* molarity

= 1*0.845 = 0.845N

N = NBVB-NAVA/VA+VB

= 0.25*253.5- 0.845*75/253.5+75 =0

no of moles of Ba(OH)2 = no of moles of HCl

so it is neutral solution

PH = 7

queen_honey_blossom answered 1 year ago

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