NaOH(aq) + HBr(aq) → H2O(l) + NaBr(aq)
25.0 mL of a 1.02M HBr solution contains "n(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)=25.5 mmol"
Volume of NaOH solution can be calculated:
"V(NaOH)=\\frac{n(NaOH)}{c(NaOH)}=\\frac{25.5mmol}{0.653\\frac{mmol}{mL}}=39.1 mL"
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