Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: integral0 on November 09, 2004, 10:39:52 PM
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Calculate the concentration (M) of arsenic acid (H3AsO4) in a solution if 25.00mL of that solution required 35.21 mL of 0.1894 M KOH for neutralization.
The answer is .08892
Any help appreciated :)
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First, calculate the no. moles KOH used
n=cv = 0.1894M x 0.03521L
=0.006669moles of KOH
Seeings 3 OH's are required to neutralise every H3AsO4
Moles of H3AsO4 = Moles KOH /3
=0.002223 moles
Now C =n/v =0.002223/0.02500
=0.08892M