50.00 mL of barium hydroxide solution is titrated to completion with 42.87 mL of 0.08713 M...

Question

Question

50.00 mL of barium hydroxide solution is titrated to completion with 42.87 mL of 0.08713 M HI. What is the M of the base?

science chemistry

Answer 1

Answer #1

Ba(OH)2 (aq) + 2HI(aq) --------------------> BaI2(aq) + 2H2O(l)

1 mole 2moles

Ba(OH)2 HI

M1 = M2 = 0.08713M

V1 = 50ml V2 = 42.87ml

n1 = 1 n2 = 2

M1V1/n1 = M2V2/n2

M1 = M2V2n1/V1n2

= 0.08713*42.87*1/(50*2)

= 0.03735M

The molarity of Ba(OH)2 = 0.03735M >>>>answer

Answer 2

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