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  1. Problems
  2. /0.360 g of a pure carboxylic acid needs 30.0 mL of sodium hydroxide solution of 0.200 mol/L for…
MediumChemistryHigh School (11–12)Organic Compounds with Functional Groups › Aldehydes, Ketones and Carboxylic AcidsNumerical

0.360 g of a pure carboxylic acid needs 30.0 mL of sodium hydroxide solution of 0.200 mol/L for…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

0.360 g0.360\ \text{g}0.360 g of a pure carboxylic acid needs 30.0 mL30.0\ \text{mL}30.0 mL of sodium hydroxide solution of 0.200 mol/L0.200\ \text{mol/L}0.200 mol/L for complete neutralisation. Find the amount of sodium hydroxide, the molar mass of the acid if it is monobasic, and the molar mass if it is dibasic. Name the acid if it is monobasic.

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

mmol
g/mol
g/mol
Need a reminder? The key ideaShowHide

Ka=10−pKaK_a = 10^{-\text{p}K_a}Ka​=10−pKa​. For a weak acid HA at concentration ccc, Ka=x2c−xK_a = \dfrac{x^2}{c - x}Ka​=c−xx2​, so x=[H+]=−Ka+Ka2+4Kac2x = [\text{H}^+] = \dfrac{-K_a + \sqrt{K_a^2 + 4K_a c}}{2}x=[H+]=2−Ka​+Ka2​+4Ka​c​​, and pH=−log⁡x\text{pH} = -\log xpH=−logx. The approximation x≈Kacx \approx \sqrt{K_a c}x≈Ka​c​ holds only when the acid is ionised by a few per cent. Withdrawing groups make an acid stronger, and the effect falls with distance. In a titration, moles of base =c×V= c \times V=c×V.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. n(NaOH)=0.200 mol/L×0.0300 L=6.00 mmoln(\text{NaOH}) = 0.200\ \text{mol/L} \times 0.0300\ \text{L} = 6.00\ \text{mmol}n(NaOH)=0.200 mol/L×0.0300 L=6.00 mmol.

Step 2. Monobasic. The acid is 6.00 mmol6.00\ \text{mmol}6.00 mmol, so M=0.360 g0.00600 mol=60.0 g/molM = \dfrac{0.360\ \text{g}}{0.00600\ \text{mol}} = 60.0\ \text{g/mol}M=0.00600 mol0.360 g​=60.0 g/mol. This is ethanoic acid.

Step 3. Dibasic. The acid is 3.00 mmol3.00\ \text{mmol}3.00 mmol, so M=0.3600.00300=120 g/molM = \dfrac{0.360}{0.00300} = 120\ \text{g/mol}M=0.003000.360​=120 g/mol.

Answer: 6.00 mmol6.00\ \text{mmol}6.00 mmol, 60.0 g/mol60.0\ \text{g/mol}60.0 g/mol and 120 g/mol120\ \text{g/mol}120 g/mol. If the acid is monobasic, it is ethanoic acid.

Check. The molar mass of ethanoic acid is 60.05 g/mol60.05\ \text{g/mol}60.05 g/mol, which agrees.

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  1. Acidity of carboxylic acidsCarboxylic Acids: Structure, Acidity and Reactions

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