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# Concentration, titration, pH and pOH of solutions.

Please see the attached file for the fully formatted problems.

Problem 10.53
What is the molarity of an HCl solution if 11.0 mL HCl solution is titrated with 21.6 mL of 0.130M NaOH solution?
HCl (aq) + NaOH (aq) &#61664; NaCl (aq) + H2O (l)

Problem 10.85
What are the [H3O+] and [OH-] for a solution with the following pH values?

Part A
pH = 3.60

Part B
pH = 6.56

Part C
pH = 8.65

Part D
pH = 11.40

Problem 10.87
What is the OH- in a solution that contains 0.220 g NaOH in 0.265 L of solution?

Problem 10.89
What is the pH of a solution prepared by dissolving 4.5 g HCl in water to make 415 mL of solution?

Problem 10.103
There is a solution of 6.6×10&#8722;2 M KOH.
Part A
Determine [H3O+].

Part B
Determine pH.

Part C
Determine products when reacted with excess of H3PO4.

Part D
Determine the number of milliliters required to neutralize 54.0 mL of 7.0×10&#8722;2 M H2SO4

Problem 10.25
What is the conjugate base of H2SO3?

Problem 10.29
Why are the concentrations of [H3O+] and [OH-] equal in pure water?

#### Solution Preview

I deleted your masteringchemistry links so they wouldn't mess up anything on my end. The answers are formatted in the Word document attached as well as pasted below.

What is the molarity of an HCl solution if 11.0 mL HCl solution is titrated with 21.6 mL of 0.130M NaOH solution?
HCl (aq) + NaOH (aq) &#61664; NaCl (aq) + H2O (l)

Since the acid and base react 1:1, you need to find out how many moles of base reacted, which you can do since you know the volume and concentration. First, convert mL to L by using the conversion factor 1000mL/L, then multiply by the molarity.
(.0216L NaOH)(.130 mol/L NaOH) = .002808 mol NaOH = .002808 mol HCl
Then divide mol HCl by L HCl to get the molarity.
(.002808 mol HCl)/(.011L HCl) = .2552727 mol/L = 0.255M

Problem 10.85
What are the [H3O+] and [OH-] for a solution with the following pH values?

Part A
pH = 3.60

Since pH = -log[H+] and Kw = 1.0 x10^-14 = [H+][OH-] you can find both. Note that [H+] is just shorthand for [H3O+].
3.60 = -log[H+] so the inverse log of -3.60 = [H+] = 2.512x10^-4
[OH-]=Kw/[H+] by rearranging the above = 3.981x10^-11
You can double check your answer in that pH + pOH = 14, and since the negative log of the above = 10.4, you're right.

Part B
pH = 6.56