Two solution A and B, each of 100 L was made by dissolving 4g of NaOH and 9.8g of H2SO4 in water, resp. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is:
( JEE MAINS - 2020 )
( Solve it including the required formula and explanation )
Answers
Answer:
10.5
Explanation:
For given solutions, we have:
Moles of NaOH 4/40 = 0.1 moles
Moles of H2SO4 = 9.8/98 = 0.1 moles
Molarity of NaOH = 0.1/100L
And molarity of H2SO4 = 0.1/100
Now, 40L of NaOH solution and 10L of H2SO4 solution are added, thus we get:
Total volume = 50L
Milliequivalents of NaOH = 40x(0.1/100)x1 = 0.04
Milliequivalents of H2SO4 = 10x(0.2/100)x2 = 0.02
Thus, Meq of NaOH left = 0.04 - 0.02 = 0.02
[OH-] = 4x10-4
pOH = -log[4x10-4]
pOH = -log4 - log10-4
pOH = -0.60 + 4 = 3.4
Further, we know:
pH = 14 - 3.4
pH = 10.5
The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is 10.6
Moles of NaOH 4/40 = 0.1 moles
Moles of H2SO4 = 9.8/98 = 0.1 moles
Molarity of NaOH = 0.1/100
Molarity of H2SO4 = 0.1/100
40L of NaOH solution and 10L of H2SO4 solution are added (Given)
∴ Total volume = 50L
Milliequivalents of NaOH = 40 x (0.1/100) x 1 = 0.04
Milliequivalents of H2SO4 = 10 x (0.2/100) x 2 = 0.02
∴ Milliequivalents of NaOH left = 0.04 - 0.02 = 0.02
[OH-] = 4x10-4
pOH = -log [4x10-4]
pOH = -log4 - log10-4
pOH = -0.60 + 4 = 3.4
pH = 14 - 3.4
pH = 10.6