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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