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.

Respuesta :

Answer:

0.0006 mole

Explanation:

Considering:

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

Or,

[tex]Moles =Molarity \times {Volume\ of\ the\ solution}[/tex]

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 :

[tex]Moles=0.100 \times {6.0\times 10^{-3}}\ moles[/tex]

Moles of HCl = 0.0006 moles

From the reaction shown below:-

[tex]HCl+Na_2CO_3\rightarrow NaHCO_3+NaCl[/tex]

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

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