A 1 mL sample of dilute acetic acid solution required 4 mL of .2M NaOH solution for complete neutralization. The density of the HC2H3O2 solution was 1.05g/mL. Calculate the percentage of HC2H3O2 in the solution.Calculate the percentage of HC2H3O2 in the solution?
Here it is in excruciating detail..........
CH3COOH (as it's normally written) is a monobasic acid, so 1 mole will be neutralized by one mole of NaOH
CH3COOH + NaOH =%26gt; CH3COONa + H2O
0.2M NaOH solution contains 0.2 moles/liter (always write the leading zero),
so 4 mL contains 0.2 x 4/1000 = 0.8/1000moles. This must also be the number of moles of CH3COOH in 1 ml of the acetic acid solution.
Therefore 1 L of the acetic acid solution will contain 0.8 moles of CH3COOH and its molarity is 0.8M
0.8 moles of CH3COOH = 0.8 x (24 + 4 + 32) = 48 g of CH3COOH per liter
Solution density is 1050 g/L
Weight % of CH3COOH is 100 x 48/1050 = 4.57%
You can of course keep the calculations on a per ml basis. I switched to liters to avoid decimals
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