Question #21145

What is the concentration of HCL, when 50.0 mL of the solution was neutralized by 12 mL of a 0.35 mol/L solution of KOH?

Expert's answer

For answering this question you need to use next law:

N_{1}V_{1} = N_{2}V_{2},

where N is normality (in this case it is the same that molaritiy) and V is volume,

So N_{2} = N_{1}V_{1}/V_{2} = 12*0.35 / 50.0 = 0.084 mol/l

N

where N is normality (in this case it is the same that molaritiy) and V is volume,

So N

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