[integral0]

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

[Tetrahedrite]

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