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5.0 cm3 of 0.050 mol/dm3 sodium carbonate was completely neutralised by 20.00 cm3 of dilute hydrochloric acid. Calculate the concentration in mol/dm3 of the hydrochloric acid.

Question

5.0 cm3 of 0.050 mol/dm3 sodium carbonate was completely neutralised by 20.00 cm3 of dilute hydrochloric acid. Calculate the concentration in mol/dm3 of the hydrochloric acid.

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Solution

Sure, here's how you can calculate the concentration of the hydrochloric acid:

Step 1: Calculate the number of moles of sodium carbonate (Na2CO3).

The number of moles = concentration x volume. So, the number of moles of Na2CO3 = 0.050 mol/dm3 x 5.0 cm3 x (1 dm3/1000 cm3) = 0.00025 mol.

Step 2: Write the balanced chemical equation for the reaction.

Na2CO3 + 2HCl -> 2NaCl + H2O + CO2

From the balanced equation, you can see that 1 mole of Na2CO3 reacts with 2 moles of HCl.

Step 3: Calculate the number of moles of HCl.

So, the number of moles of HCl = 2 x number of moles of Na2CO3 = 2 x 0.00025 mol = 0.0005 mol.

Step 4: Calculate the concentration of HCl.

The concentration = number of moles/volume. So, the concentration of HCl = 0.0005 mol / 20.00 cm3 x (1 dm3/1000 cm3) = 0.025 mol/dm3.

So, the concentration of the hydrochloric acid is 0.025 mol/dm3.

This problem has been solved

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