Question

what will be the nature of the resultant solution when 20 ml, 0.1m H2SO4 is mixed with 20 ml, 0.1m Ba(OH) 2 solution​

Answer 1

Answer:

First write down the balanced equation reaction :

H2SO4 + 2 NaOH → Na2SO4 + 2 H2O

The reaction shows that 1 mole of H2SO4 reacts with 2 moles of NaOH → the mole ratio between H2SO4 and NaOH is 1 : 2. The total volume of the solution is 26 mL.

Then calculate the mole of both H2SO4 and NaOH. Here, the mole of H2SO4 = 20 mL x 0.1 M = 2 mmoles and the mole of NaOH = 6 mL x 0.025 M = 0.15 mmole. In this case, NaOH acts as the limiting reagent.

So, 0.15 mmole of NaOH reacts with 1/2 x 0.15 mmole of H2SO4 or 0.075 mmole of H2SO4.

Since the initial mole of H2SO4 being 2 mmoles, and of this amount is consumed by NaOH as much as 0.075 mmole, thus the amount of H2SO4 left over = 2 mmoles - 0.075 mmole = 1.925 mmoles.

Thus, the concentration of H2SO4 in the solution is 1.925 mmoles/26 mL. One can expressed this in 1 L (1000 mL), so the concentration of H2SO4 is 1000/26 x 1.925 mmoles/L or 74.038 mmoles/L or (74.038 mmoles x mole/1000 mmoles)/L of H2SO4 solution or 0.074038 mole/L of H2SO4 solution.

Here, the molarity of H2SO4 has been 0.074038 M→ rounded up to 0.074 M, or 7.4 x 10^-3 M.

pH = -log [H+] → pH = -log [7.4 x 10^-3]

pH = 2.13