Question

# CHEMISTRY! A solution of a triprotic acid is prepared by dissolving 6.088 g of the solid...

CHEMISTRY!

A solution of a triprotic acid is prepared by dissolving 6.088 g of the solid acid in sufficient DI water to make 400.00 mL of solution.   18.36 mL of a 0.1745 M NaOH solution are required to neutralize 10.00 mL of this acid solution?

A) What is the concentration of the acid solution?

B) What is the molar mass of the acid?

NOTE: this was the response to my answer -> .961146

This would be correct for a monoprotic acid titrated with a triprotic base. In this question, a triprotic acid is titrated with a monoprotic base.

Reaction between a triprotic acid and NaOH will be

H3A + 3NaOH Na3A + 3H2O

So the total moles of NaOH used in this titration will be 3 times the number of moles of the triprotic acid

number of moles of NaOH is 18.36 mL of a 0.1745 M NaOH = 0.01836 x 0.1745 = 0.00320 moles of NaOH is required to neutralize 10 mL of the acid

so for 400 mL of the acid we will need 0.00320 x 40 = 0.128 moles

since 0.128 moles is in 400 mL the concentration is (1000 x 0.128)/400 = 0.32 M solution

If 0.128 moles is 6.088 g the 1 mole will be 6.088/0.128 = 47.56 is the molar mass of the acid.

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