Posted by nelson on Monday, March 31, 2008 at 7:43am.
Convert 0.140 g KOH to mols. mols = grams/molar mass.
There is 1 mol K^+ per mol KOH.
Then [K^+] = mols/0.250 L = ??
Let's designate, for simplicity in typing, aniline as RNH2.
RNH2 + HOH ==> RNH3^+ + OH^-
Kb = (RNH3^+)(OH^-)/(RNH2) = 7.4 x 10^-10
(RNH3^+) = x
(OH^-) = x
(RNH2) = 0.570 - x
Substitute into Kb and solve for x = (OH^-), convert to (H^+) and solve for pH. Post your work if you get stuck.
Check my thinking.
Related Questions
CHEM - What is the molarity of potassium iodide solution prepared by adding 5....
Chem - What is the hydroxide ion concentration of a solution prepared by ...
Chemistry - A student makes a solution by dissolving 55.8 grams of potassium ...
Chemistry - A student makes a solution by dissolving 55.8 grams of potassium ...
Chemistry - A student makes a solution by dissolving 55.8 grams of potassium ...
CHEMISTRY - Consider an aqueous solution prepared from 250.0 mL of water and 1....
chemisrty - Q4) Suppose you prepare 950 mL of a solution by dissolving 100 grams...
Chemistry - Find the pH of a solution prepared by taking a 50.0 mL aliquot of a ...
Chemistry - A student makes a solution by dissolving 55.8 grams of potassium ...
chemistry - A potassium nitrate solution is prepared by dissolving 2.3 g of KNO3...
For Further Reading