In a 0.050 M solution of a weak monoprotic acid,[H + ] = 1.8 x 10 -3 . What is i
ID: 684055 • Letter: I
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In a 0.050 M solution of a weak monoprotic acid,[H+] = 1.8 x 10-3. What is itsKa? A. 3.6 x 10-2 B. 9.0 x 10-5 C. 6.7 x 10-5 D. 1.6 x 10-7 Please EXPLAIN youranswer. In a 0.050 M solution of a weak monoprotic acid,[H+] = 1.8 x 10-3. What is itsKa? A. 3.6 x 10-2 B. 9.0 x 10-5 C. 6.7 x 10-5 D. 1.6 x 10-7 Please EXPLAIN youranswer. Please EXPLAIN youranswer.Explanation / Answer
always when doing acid/base reactions, the first thing you shoulddo is write out your reaction with the amounts Now if you haven't been taught this yet, the common way to solvethese is with an ice chart, in later years of chemistry you'lllearn some easy calculations to just solve from the informationalready given. ICE = Initial - Change - Equilibrium You start with 0.05M of HA and end up with [H+] = 1.8 x10-3 where the H+ is found in the equilibriumstate. Also water does not affect this equation. HA(aq) + H2O(l) --->A-(aq) + H+(aq) I 0.05 - - - C -x - +x +x E 0.05-x - x x In this case, x = 1.8 x 10-3 therefore you can solvethis out for the final concentrations of all your species. Now the standard Ka expression is: Ka =[Products]/[Reactants] -brackets meaningconcentrations You have all the concetrations from the ICE chart calculations soall that is left is to plug in and solve. Ka = [A-][H+]/[HA] = (1.8 x10-3)(1.8 x 10-3) / (0.0482) Ka = 6.72 x 10-5 this seems like a reasonable Ka for a weak acid and to check it youcan take the pKa which is the -log(ka) and see where it lies on theacidity chart. -log(6.72 x 10-5) = 4.17 which is below pH 7 thereforeshowing it is an acid and since it is above a pH of 2-3 it isweakly acidic. If you need anymore help with this please follow up on it. ICE tables can be confusing at first but it really just takestime. After you do enough practice you can do them quickly oreven learn the patterns to skip them.
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