(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 3.125 g sample of primary standard of Na2CO3 was treated with 40.00 mL of dilute perchloric acid. The solution was boiled to remove CO2, following which the excess HClO4 was back-titrated with 10.12 mL of dilute NaOH. In a separate experiment, it was established that 27.43 mL of the HClO4 neutralized the NaOH in a 25.00 mL portion. Calculate the molarity of the NaOH solution. Hint: Solve this problem by considering the two equations below

2. Relevant equations

For the back-titration (1): total amount of perchloric acid: n1(HClO4) = 2n(Na2CO3) + n1(NaOH)

For the acid-base titration (2): n2(NaOH) =n2(HClO4)

3. The attempt at a solution

I first calculate the moles for equation (1)

(40/1000)[HClO4]=2*(3.125g/ (105.99g/mol) +(10.12/1000)[NaOH]

then eq. (2)

(27.43/1000)[HClO4]=(25/1000)[NaOH]

I then substituted equation 2 into equation 1. My final answer for [NaOH] was 2.239. However the actual answer was 1.974. Can somebody please point out what I did wrong. Thanks

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Calculate the molarity of a NaOH solution

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