Imagine you had HCl with a concentration of exactly 0.10 mol/dm3. If 0.023 dm3 of a sodium hydroxide solution, NaOH (aq), could exactly neutralize 0.040 dm3 of the HCl solution, what is the concentration of the NaOH (aq)

Answer:

Explanation:

Step 1: Calculate the amount of sodium hydroxide in moles

Volume of sodium hydroxide solution = 25.0 ÷ 1,000 = 0.0250 dm3

Rearrange:

Concentration in mol/dm3 =

Amount of solutein mol = concentration in mol/dm3 × volume in dm3

Amount of sodium hydroxide = 0.100 × 0.0250

= 0.00250 mol

Step 2: Find the amount of hydrochloric acid in moles

The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

So the mole ratio NaOH:HCl is 1:1

Therefore 0.00250 mol of NaOH reacts with 0.00250 mol of HCl

Step 3: Calculate the concentration of hydrochloric acid in mol/dm3

Volume of hydrochloric acid = 20.00 ÷ 1000 = 0.0200 dm3

Concentration in mol/dm3 =

Concentration in mol/dm3 =

= 0.125 mol/dm3

Step 4: Calculate the concentration of hydrochloric acid in g/dm3

Relative formula mass of HCl = 1 + 35.5 = 36.5

Mass = relative formula mass × amount

Mass of HCl = 36.5 × 0.125

= 4.56 g

So concentration = 4.56 g/dm3

What happens when a current is applied to two electrodes immersed in pure water?

A. No electrolysis occurs, and current does not flow.

B. Oxygen, but not hydrogen, is released as a gas.

C. Electrolysis occurs, releasing oxygen and hydrogen gas.​