you prepared the buffer by adding about 0.2 g of each of the weak acid and its c

Question

you prepared the buffer by adding about 0.2 g of each of the weak acid and its conjugate base to 25.0 ml of water. The goal was to produce a buffer near it point of maximum buffer capacity where [WA]=[CB] An alternative could have been to dissolve about 0.4 grams of the weak acid in water, and add enough NaOH to convert half the weak acid into its conjugate base. how many ml of 0.10 mol of NaOH would have been needed to achieve this result. 1 ml of NaOH was given the Delta PH was 0.113. The mass of the weak acid was 0.185 g. and the molecular weight was 234 g. The mass of the conjugate base was 0.191 g. and the molecular weight was 256g

Explanation / Answer

moles of weak acid present = 0.185 g/234 g/mol = 0.0008 mol

moles of conjugate base present = 0.191 g/256 g/mol = 0.0007 mol

So for the buffer with maximum buffer capacity,

[WA] = [CB]

we need to add moles of 0.1 M NaOH = 0.00005 mol

Volume of 0.1 M NaOH to be added for the buffer solution = 0.00005 x 1000/0.1 = 0.5 ml

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