Ionic EquilibriumHard
Question
50 mL of 0.1 M NaOH is added to 60 mL of 0.15 M H3PO4 solution (K1, K2 and K3 for H3PO4 are 10-3, 10-8 and 10-13 respectively). The pH of the mixture would be about (log 2 = 0.3)
Options
A.3.1
B.5.5
C.4.1
D.6.5
Solution
H3PO4 + NaOH →NaH2PO4
9 5 0
4 0 5
pH = pKa + log
= 3 + 0.1 = 3.1
9 5 0
4 0 5
pH = pKa + log
= 3 + 0.1 = 3.1 Create a free account to view solution
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