Answer to Question #344158 in General Chemistry for Meg

Question #344158

During a titration, 0.200 M HCl is added to a NaOH solution of unknown concentration. What is the concentration of the NaOH solution if 20.0mL of it is neutralized by 30.7mL of the standard solution? Show your work.


1
Expert's answer
2022-05-25T10:46:03-0400

The amount of mole at the neutralization point is equal for both components of the reaction:

n1=n2n_1=n_2

cNaOHVNaOH=cHClVHClc_{NaOH}V_{NaOH}=c_{HCl}V_{HCl}

cNaOH=cHClVHClVNaOH=0.200mol/L×30.7ml20.0ml=0.307mol/Lc_{NaOH} = \frac{c_{HCl}V_{HCl}}{V_{NaOH}} = \frac{0.200 mol/L\times 30.7 ml}{20.0 ml} = 0.307 mol/L

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