Asked by Anon
Calculate the pH of the solution formed when 20mL of 0.100 mol/L HNO_3 is added to 30 mL of 0.050 mol/L KOH.
Answers
Answered by
DrBob222
millimols HNO3= mL x M = 20 x 0.1 = 2.0
millimols KOH = 30 x 0.05 = 1.0
Total volume is 20 mL + 30 mL = 50 mL
.............................KOH + HNO3 ==> KNO3 + H2O
Initial......................1.0..........0...............0...........0
add......................................2.0....................................
Change.................-1.0........-1.0...........1.0............1.0
Equilibrium...............0...........1.0............1.0
So the chart above tells you that when the reaction is complete you have 1.0 millimols left of HNO3 in 50 mL. Since millimols = mL x M then M = millimols/mL = 1.0 mmols/50 mL = 0.02 M
Use pH = -log(HNO3) to convert to pH. Post your work if you get stuck.
millimols KOH = 30 x 0.05 = 1.0
Total volume is 20 mL + 30 mL = 50 mL
.............................KOH + HNO3 ==> KNO3 + H2O
Initial......................1.0..........0...............0...........0
add......................................2.0....................................
Change.................-1.0........-1.0...........1.0............1.0
Equilibrium...............0...........1.0............1.0
So the chart above tells you that when the reaction is complete you have 1.0 millimols left of HNO3 in 50 mL. Since millimols = mL x M then M = millimols/mL = 1.0 mmols/50 mL = 0.02 M
Use pH = -log(HNO3) to convert to pH. Post your work if you get stuck.
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