A titration study of the following reaction was performed: 2 NaOH(aq) + H­ 2 SO

Question

A titration study of the following reaction was performed:

            2 NaOH(aq) + H­2SO4(aq) Na2SO4(aq) + 2 H2O

When 8.35mL of 0.1385M NaOH solution from a buret was added to 25.00mL of 0.0986M H2SO4, in a flask, all the NaOH was consumed by reaction with some of the H2SO4. Determine the number of moles of unreacted H2SO4remaining in the flask at this stage of the titration.

*** Show your work

***The answer is 1.89 x 10^ -3 mol...but PLEASE show all of your work CLEARLY so I can understand it! THANK YOU!!

Explanation / Answer

Initial H2SO4 moles = M x V ( in L) = 0.0986 x ( 25/1000) = 0.002465

NaOH moles = Mx V of NaOH = ( 0.1385 ) x ( 8.35/1000) = 0.001156475

as per reaction 2 NaOH reacts per 1 H2SO4

hence H2SO4 moles reacted with 0.001156475 moles NaOH = (0.001156475/2) = 0.0005782375

now excess H2SO4 left = H2SO4 moles initial - H2SO4 moels reacted with NaOH

       = 0.002465 - 0.0005782375 = 0.00189 = 1.89 x 10^ -3 mol

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