Answer to Question #245269 in General Chemistry for Sabra

Question #245269

Calculate the pH of a solution prepared by mixing 2 ml of a strong acid solution of pH 3 and 3 ml of a strong base of pH 10


1
Expert's answer
2021-10-07T03:45:08-0400

pH of Strong Acid = 3

-log[H+] = 3

[H+] = 10,-3


pH of strong Base = 10

\therefore pOH of strong base = 14 - 10 = 4

log[OH-] = 4

[OH-] = 10-4


Moles of H+ in 2 ml solution = 2 × 10-3 × 10-3

Moles of OH- in 3 ml solution = 3 × 10-4 × 10-3

Now, Reaction

H^++OH^- \quad^{\rightarrow }_\leftarrow \quad H_2O_{(l)}


Moles of H+ remaining in (3+2)ml solution = (2*l×10-6 - 3×10-7) = 1.7 x 10-6


Molarity of H+ in solution = 1.7×1065×1000=3.40×104\dfrac{1.7×10^{-6}}5×1000 = 3.40× 10^{-4}


[H+] = 3·40 x 10-4

pH = -log[H+]

pH = -log [3-40х10-4]

pH = 3.46 \approx 3.5


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