25.49 ml of NaOH were required to neutralize .5208g of KHP (m.w.204.33) dissolve

Question

25.49 ml of NaOH were required to neutralize .5208g of KHP (m.w.204.33) dissolved in water. This titration served to standardizethe NaOH. The vinegar solution for titration was prepared inthe following manner: 25 ml of vinegar were diluted to 250 ml in avolumetric flask and 25 ml of this diluted solution required 22.62ml of the above standardized NaOH to reach the phenolphthaleinendpoint.

1) What was the molarity of the NaOH solution used?
2) What is the percentage by mass of the acid in the originalvinegar sample?

Explanation / Answer

(a) Since the ration is 1:1 0.5208 g KHP * 1 mol KHP/ 204.23 g * 1mol NaOH / 1 mol KHP =0.00255 mol NaOH If 22.62 ml were used and we know M = mole/L M = 0.00255 mol / 25.49* 10^-3 L = 0.1000 M is the molarity ofNaOH solution (b) by acid you mean acetic acid. If the rection is one toone, and we now the molarity of the solution 22.62*10^-3 L * 0.1000mol/L *1 mol AA/1 mol NaOH * 60.05 g AA/1 mol = 0.1358 g acetic acid divide the answer by the total mass of the vinegar andmultiply times 100 to get the solution

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