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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: jay426 on November 11, 2008, 10:50:02 am

Title: questions from IARTV exam.
Post by: jay426 on November 11, 2008, 10:50:02 am: A solution of limewater contains 5x10^-5 mol of Ca(OH)2 in 10mL of the solution at 25C. The pH of this solution is? A. 2 B. 2.30 C. 11.7 D. 12

Answer says pH=12. How do you get 12?
Title: Re: Quick question from IARTV exam.
Post by: Collin Li on November 11, 2008, 10:53:01 am: Since: $\mbox{Ca(OH)}_2 \rightarrow \mbox{Ca}^{2+} + 2\mbox{OH}^-$

$\mbox{[OH}^-\mbox{]} = \frac{n}{V} = \frac{2\times\left(5\times10^{-5}\right)}{0.010} = 0.01\mbox{ M}$

Since $\mbox{[H}^+\mbox{]}\mbox{[OH}^-\mbox{]} = 10^{-14}$

$\implies \mbox{[H}^+\mbox{]} = \frac{10^{-14}}{0.01} = 10^{-12}\mbox{ M}$

$\implies \mbox{pH} = -\log_{10} \left(10^{-12}\right) = 12$
Title: Re: Quick question from IARTV exam.
Post by: jay426 on November 11, 2008, 10:58:46 am: thanks. that makes sense now.
another question though that i can't get. A 10mL solution of hypobromous acid has a pH of 5. what is the concentration of hypobromous acid? answer is .042 mol L^-1
Title: Re: Quick question from IARTV exam.
Post by: Collin Li on November 11, 2008, 11:19:55 am: No idea what that is, so your first instinct should be to check your data book!

And it tells you it has a $K_a = 2.4\times 10^{-9}$

So, since we know $\mbox{[H}^+\mbox{]} = 10^{-5}\mbox{ M}$ , and for any acid:

$\mbox{HA} \rightarrow \mbox{H}^+ + \mbox{A}^-$

(In this case, for hypobromous acid, $\mbox{A}^- = \mbox{OBr}^-$ )

Then, $\mbox{[A}^-\mbox{]} = \mbox{[H}^+\mbox{]} = 10^{-5}\mbox{ M}$

So: $K_a = \frac{\mbox{[H}^+\mbox{]}\mbox{[A}^-\mbox{]}}{\mbox{[HA]}} = \frac{\mbox{[H}^+\mbox{]}^2}{\mbox{[HA]}}$

Substituting in values yields:

$2.4 \times 10^{-9} = \frac{10^{-10}}{\mbox{[HA]}}$

$\implies \mbox{[HA]}_e = \frac{10^{-10}}{2.4 \times 10^{-9}} = 0.042\mbox{ M}$

You can stop here, but you might lose a mark if you don't include this subtle step:

Since hypobromous acid is a weak acid, we can assume $\mbox{[HA]}_i \approx \mbox{[HA]}_e$

$\therefore \mbox{[HA]}_i = 0.042\mbox{ M}$

The reason why we need to make that last step is because we only solved for the equilibrium amount of hypobromous acid. Now we need to find out the concentration of the hypobromous acid initially in the bottle. In acidity constant calculations involving weak acids, we always have a step similar to this.
Title: Re: questions from IARTV exam.
Post by: jay426 on November 11, 2008, 11:39:32 am: thanks. Also, lol another question. Three electrolysis cells are connected in series, each one producing a different metal. The metals produced are aluminium, magnesium and sodium. After one hour with a constant applied current, which cell deposits the greatest amount of metal? answer is sodium.
Title: Re: questions from IARTV exam.
Post by: Collin Li on November 11, 2008, 12:00:15 pm: This is Faraday's second law.

The amount of metal deposited is inversely proportional to the size of its charge (as a metal cation).

This is because aluminium requires 3 electrons, magnesium 2, and sodium only 1. So if there is the same amount of electrons going through (same charge, or same current and time, as $Q=It$ ), then there will be more sodium than magnesium, and more magnesium than aluminium.