Two solutions A and B each of 100L was made by dissolving 4g of NaOH and 9.8g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40L of solution A and 10L of solution B is_________:
Option: 1 pH = 10.6
Option: 2 pH = 12.7
Option: 3 pH = 9.8
Option: 4 pH = 11
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 = 10.6
Hence, the option number (1) is correct.
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