A 0.100 mole quantity of a monoprotic acid HA is added to 1.ooL of pure water. W

Question

A 0.100 mole quantity of a monoprotic acid HA is added to 1.ooL of pure water. When equilibrium is reached, the pH of thesolution is 3.75. What is the value of Ka for theacid HA? This is a lab question but we have not even gotten to thischapter in class yet. Any help would be great. Promise to rate.Thanks. K8 A 0.100 mole quantity of a monoprotic acid HA is added to 1.ooL of pure water. When equilibrium is reached, the pH of thesolution is 3.75. What is the value of Ka for theacid HA? This is a lab question but we have not even gotten to thischapter in class yet. Any help would be great. Promise to rate.Thanks. K8

Explanation / Answer

c(acid) = n/v = 0.1/1 = 0.1M pH = 3.75 pH = -log [H+] log [H+] = -3.75 [H+] = 0.0001778M [H+] = (KaCa) 0.00017782 = Ka x 0.1 Ka = 3.16 x 10-7 Hope this helps!

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