Question

Calculate the number of moles of sodium carbonate originally present, if 6.0 ml of 0.100 M HCl were used to titrate the mixture with phenolphthalein as the indicator. Show your calculations.

Answer 1

Answer:

0.0006 mole

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For HCl :

Molarity = 0.100 M

Volume = 6.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 6.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=0.100 \times {6.0\times 10^{-3}}\ moles$

Moles of HCl = 0.0006 moles

From the reaction shown below:-

$HCl+Na_2CO_3\rightarrow NaHCO_3+NaCl$

1 mole of HCl reacts with 1 mole of sodium carbonate.

So,

0.0006 mole of HCl reacts with 0.0006 mole of sodium carbonate.

Moles of sodium carbonate = 0.0006 moles