# Acid-base reactions  (Page 4/5)

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When you are busy with these calculations, you will need to remember the following:

1dm ${}^{3}$ = 1 litre = 1000ml = 1000cm ${}^{3}$ , therefore dividing cm ${}^{3}$ by 1000 will give you an answer in dm ${}^{3}$ .

Some other terms and equations which will be useful to remember are shown below:

• Molarity is a term used to describe the concentration of a solution, and is measured in mol.dm ${}^{-3}$ . The symbol for molarity is M. Refer to chapter [link] for more information on molarity.
• Moles = molarity (mol.dm ${}^{-3}$ ) x volume (dm ${}^{3}$ )
• Molarity (mol.dm ${}^{-3}$ ) = $\frac{moles}{volume}$

Given the equation:

$\mathrm{NaOH}+\mathrm{HCl}\to \mathrm{NaCl}+{\mathrm{H}}_{2}\mathrm{O}$

25cm ${}^{3}$ of a sodium hydroxide solution was pipetted into a conical flask and titrated with 0.2M hydrochloric acid. Using a suitable indicator, it was found that 15cm ${}^{3}$ of acid was needed to neutralise the alkali. Calculate the molarity of the sodium hydroxide.

1. NaOH: V = 25 cm ${}^{3}$

HCl: V = 15 cm ${}^{3}$ ; C = 0.2 M

The equation is already balanced.

2. $M=\frac{n}{V}$

Therefore, n(HCl) = M $×$ V (make sure that all the units are correct!)

M = 0.2mol.dm ${}^{-3}$

V = 15cm ${}^{3}$ = 0.015dm ${}^{3}$

Therefore

$n\left(HCl\right)=0.2×0.015=0.003$

There are 0.003 moles of HCl that react

3. Look at the equation for the reaction. For every mole of HCl there is one mole of NaOH that is involved in the reaction. Therefore, if 0.003 moles of HCl react, we can conclude that the same quantity of NaOH is needed for the reaction. The number of moles of NaOH in the reaction is 0.003.

4. First convert the volume into dm ${}^{3}$ . V = 0.025 dm ${}^{3}$ . Then continue with the calculation.

$M=\frac{n}{V}=\frac{0.003}{0.025}=0.12$

The molarity of the NaOH solution is 0.12 mol.dm ${}^{3}$ or 0.12 M

4.9 g of sulfuric acid is dissolved in water and the final solution has a volume of 220 cm ${}^{3}$ . Using titration, it was found that 20 cm ${}^{3}$ of this solution was able to completely neutralise 10 cm ${}^{3}$ of a sodium hydroxide solution. Calculate the concentration of the sodium hydroxide in mol.dm ${}^{-3}$ .

1. ${\mathrm{H}}_{2}{\mathrm{SO}}_{4}+2\mathrm{NaOH}\to {\mathrm{Na}}_{2}{\mathrm{SO}}_{4}+2{\mathrm{H}}_{2}\mathrm{O}$

2. M = n/V

V = 220 cm ${}^{3}$ = 0.22 dm ${}^{3}$

$n=\frac{m}{M}=\frac{4.9g}{98g.mo{l}^{-1}}=0.05mols$

Therefore,

$M=\frac{0.05}{0.22}=0.23mol.d{m}^{-3}$
3. Remember that only 20 cm ${}^{3}$ of the sulfuric acid solution is used.

M = n/V, therefore n = M $×$ V

$n=0.23×0.02=0.0046mol$
4. According to the balanced chemical equation, the mole ratio of H ${}_{2}$ SO ${}_{4}$ to NaOH is 1:2. Therefore, the number of moles of NaOH that are neutralised is 0.0046 $×$ 2 = 0.0092 mols.

5. $M=\frac{n}{V}=\frac{0.0092}{0.01}=0.92M$

## Demonstration : the reaction of acids with carbonates

Apparatus and materials:

Small amounts of sodium carbonate and calcium carbonate (both in powder form); hydrochloric acid and sulfuric acid; retort stand; two test tubes; two rubber stoppers for the test tubes; a delivery tube; lime water. The demonstration should be set up as shown below.

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