Question #90010

How many milliliters of 0.653 M NaOH are needed to neutralize 25.0 mL of a 1.02 M HBr solution? The neutralization reaction is: NaOH(aq) + HBr(aq) → H2O(l) + NaBr(aq)

Expert's answer

NaOH(aq) + HBr(aq) → H2O(l) + NaBr(aq)


25.0 mL of a 1.02M HBr solution contains n(HBr)=c(HBr)⋅V(HBr)=1.02mmolmL⋅25.0mL=25.5mmoln(HBr)=c(HBr)\cdot V(HBr) = 1.02\frac{mmol}{mL}\cdot 25.0 mL = 25.5 mmol HBr. According to stoichiometry of the reaction the same amount of NaOH is required for neutralisation of HBr:

n(NaOH)=n(HBr)=25.5mmoln(NaOH)=n(HBr)=25.5 mmol

Volume of NaOH solution can be calculated:

V(NaOH)=n(NaOH)c(NaOH)=25.5mmol0.653mmolmL=39.1mLV(NaOH)=\frac{n(NaOH)}{c(NaOH)}=\frac{25.5mmol}{0.653\frac{mmol}{mL}}=39.1 mL


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