1. Calculate the anticipated pH for each of the solutions that resulted when you
ID: 824671 • Letter: 1
Question
1. Calculate the anticipated pH for each of the solutions that resulted when you added HCl or NaOH to the "water" beaker (a beaker containing 50 mL of DI water and chemical indicators).
2. Calculate the concentration of H3O+ ion for each of the solutions. Enter the data in the tables.
Here is the first row of each table, after I see how it's done, I can do the rest.
Volume of HCl added | Anticipated pH | Experimentally determined pH | Experimentally determined [H3O+]
5ml | ??? | 2.24 | ??? |
Volume of NaOH added | Anticipated pH | Experimentally determined pH | Experimentally determined [H3O+]
5ml | ??? | 11.45 | ??? |
Given:
Kw is [H3O+][OH-] = 1.0 x 10^-14
pH = -log[H3O+]
HCl concentration and NaOH concentration: 0.01M
Please show steps, any help is appreciated! :)
Explanation / Answer
Volume of HCl added = 5 mL
Moles of HCl added = volume x concentration
= 5/1000 x 0.01 = 5 x 10^(-5) mol
Total volume = 50 + 5 = 55 mL = 0.055 L
Anticipated [H3O+] = moles of HCl/total volume
= 5 x 10^(-5)/0.055 = 9.091 x 10^(-4) M
Anticipated pH = -log[H3O+] = -log(9.091 x 10^(-4)) = 3.04
Experimentally determined [H3O+] = 10^(-experimentally determined pH)
= 10^(-2.24)
= 5.754 x 10^(-3) M = 5.8 x 10^(-3) M
Volume of NaOH added = 5 mL
Moles of NaOH added = volume x concentration
= 5/1000 x 0.01 = 5 x 10^(-5) mol
Total volume = 50 + 5 = 55 mL = 0.055 L
Anticipated [OH-] = moles of NaOH/total volume
= 5 x 10^(-5)/0.055 = 9.091 x 10^(-4) M
Anticipated pOH = -log[OH] = -log(9.091 x 10^(-4)) = 3.04
Anticipated pH = 14 - anticipated pOH = 14 - 3.04 = 10.96
Experimentally determined [H3O+] = 10^(-experimentally determined pH)
= 10^(-11.45)
= 3.548 x 10^(-12) M = 3.5 x 10^(-12) M
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