Question

What volume (in mL ) of 0.200   M NaOH do we need to titrate 40.00 mL of 0.140   M HBr to the equivalence point?

Answer 1

NaOH + HBr =⇒ NaBr + H2O

35.0 ml HBr x 1 liter/1000 mL x 0.140 moles HBr/ liter = 0.0049 moles HBr

0.0049 moles HBr x 1mole NaOH/1mole HBr = 0.0049 moles HBr

0.0049 moles HBr x 1 liter NaOH/0.200 moles NaOH x 1000 mL/1liter= 24.5 mL NaOH

Answer 2

Answer: 28 ml

Explanation: $HBr+NaOH\rightarrow NaBr+H_2O$

At equivalence point the number of $H^+$ ions is equal the number of  $OH^-$ ions and thus the acid has been completely neutralized by the base.

$M_1V_1=M_2V_2$

$M_1$ = molarity of acid

$V_1$ = volume of acid

$M_2$ = molarity of base

$V_2$ = volume of base

$0.140\times 40ml=0.200\times V_2$

$V_2=28ml$

Thus 28 ml of 0.200 M NaOH are needed to titrate 40.00 mL of 0.140 M HBr to the equivalence point.