Question

Two letters of 0.2 molar H2SO4 are reacted with two electrons 0.1 molar NaOH solution. The molarity of the resulting product Na2SO4 in the solution is:

a) 0.05 M
b) 0.1 M
c) 0.2 M
d) 0.4 M

Answer 1

When two liters of 0.2 molar H2SO4 and two liters of 0.1 molar NaOH are reacted, the molarity of the resulting Na2SO4 in the solution is 0.05 M.

Step-by-step explanation:

The question presented deals with a neutralization reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH), which results in the formation of sodium sulfate (Na2SO4) and water (H2O).

Step 1: Write down the balanced chemical equation

H2SO4 (aq) + 2 NaOH (aq) → Na2SO4 (aq) + 2 H2O (l)

Step 2: Calculate the moles of NaOH

NaOH has a molarity of 0.1 M, and 2 liters of NaOH would contain 2 L × 0.1 mol/L = 0.2 mol of NaOH.

Step 3: Determine the moles of H2SO4 that reacted

According to the balanced equation, 1 mole of H2SO4 reacts with 2 moles of NaOH. So, 0.2 mol of NaOH would need 0.1 mol of H2SO4 to completely react.

Step 4: Calculate the molarity of Na2SO4 produced

Since 2 liters of solution is used, the molarity of Na2SO4 would be the moles of H2SO4 divided by 2 liters, which is 0.1 mol / 2 L = 0.05 M.