Answer:
We know that,
[H+ ] of acid solution = 1×10-3 M
Volume of acid given is 2 mL, so
Number of moles of [H+ ] = M × Vol
= 1×10-3 * 2 = 2 * 10-3 mmol
Similarly,
p[OH] of the base solution = 14 -10 = 4
[OH-] of base solution = 1 ×10-4 M
Volume of basse solution = 3 mL, so
Number of moles of [OH- ] = 3×10-4 mmol
Therefore,
Excess quality of acid = 0.002 – 0.0003 = 0.0017 mmol
[H+ ] of resulting solution = = 3.4 * 10-4 M
Thus,
pH = - log (3.4 * 10-4 ) = 4 – log 3.4 = 3.47
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