The molarity of a Sodium carbonate solution : 0.373 M
<h3>Further explanation</h3>
Given
32.52 g Na₂CO₃
822 ml of solution = 0.822 L
Required
The molarity
Solution
Molarity shows the number of moles of solute in every 1 liter of solution.

mol of solute = mol of Na₂CO₃ :
= mass : MW Na₂CO₃
= 32.52 g : 106 g/mol
= 0.307
Molarity :
= n : V
= 0.307 mol : 0.822 L
= 0.373 M
Answer:
3.336.
Explanation:
<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>
<em />
So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>
<em></em>
<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>
∵ pH = - log[H⁺].
<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>
Answer:
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