Asked by kelly
10.0mL of solution containing 20.0 mL of 1.0 M NaX (unknown base) and 10.0 mL of 1.0 M HX (unknown acid) was diluted with 50.0 ml water. The pH is 4.92. What is the concentration of HX after dilution and X- after dilution? What is the equilibrium concentration of HX, X-, and H3O+?
Answers
Answered by
DrBob222
Original solution:
moles NaX = M x L = 1.0 x 0.020 0.02 mol
moles HX = M x L = 1.0 x 0.01 = 0.01 mol
total volume = 30 mL = 0.030 L.
How much did we remove with 10 mL?
0.02 x (10/30) = 0.00667 moles NaX
0.01 x (10/30) = 0.00333 moles HX
Then took that many moles and diluted the whole thing to 50 mL (0.050 L).
Concn final solution:
M = moles/L = 0.00667/0.05 = ?? NaX
and 0.00333/0.05 = ?? HX.
This will give you NaX and HX in the final diluted solution.
That's the confusing part. The rest of the problem is easier. The equilibrium concns is done as follows:
pH is given. Convert that to H^+. That is the equilibrium H^+. X^- will be sum of H^+ + original X.
moles NaX = M x L = 1.0 x 0.020 0.02 mol
moles HX = M x L = 1.0 x 0.01 = 0.01 mol
total volume = 30 mL = 0.030 L.
How much did we remove with 10 mL?
0.02 x (10/30) = 0.00667 moles NaX
0.01 x (10/30) = 0.00333 moles HX
Then took that many moles and diluted the whole thing to 50 mL (0.050 L).
Concn final solution:
M = moles/L = 0.00667/0.05 = ?? NaX
and 0.00333/0.05 = ?? HX.
This will give you NaX and HX in the final diluted solution.
That's the confusing part. The rest of the problem is easier. The equilibrium concns is done as follows:
pH is given. Convert that to H^+. That is the equilibrium H^+. X^- will be sum of H^+ + original X.
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