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Topic: Concentration/Neutralization Question  (Read 10480 times)

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integral0

  • Guest
Concentration/Neutralization Question
« on: November 09, 2004, 10:39:52 PM »
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

  • Guest
Re:Concentration/Neutralization Question
« Reply #1 on: November 10, 2004, 12:21:51 AM »
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
« Last Edit: November 10, 2004, 12:22:34 AM by Tetrahedrite »

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