Author Topic: Ph Questions help (Read 1425 times) Tweet Share

psychlaw · « **on:** October 08, 2008, 11:52:31 pm »

I need to know how to do these
if you can solve these examples I can do the rest myself (there pretty basic questions)
thanks alot!

(1) Calculate the pH of the solution made by dissolving 3.00 g of hydrogen chloride in distilled water, and making up to 500 mL of the solution

(2) Calculate the pH of the following mixture: when 10 mL of 0.2M HCl is diluted to 20 mL of solution

(3) Determine the pH of: [NaOH] = 10^-5 M

(4) 78.0 mL of 1.28M hydropchloric acid is diluted to 750 mL with distilled water. What is the pH of the resultant solution

THANKS

Collin Li · « **Reply #1 on:** October 09, 2008, 12:08:23 am »

Question 1

Because $\mbox{HCl} \rightarrow \mbox{H}^+ + \mbox{Cl}^-$

$\mbox{[H}^+\mbox{]} = \mbox{[HCl]} = \frac{\frac{3.00}{1.0+35.5}\mbox{ mol}}{0.500\mbox{ L}} = 0.164\mbox{ M}$

$\implies \mbox{pH} = -\log_{10}0.164 = 0.784$

Question 2

Using $c_1V_1 = c_2V_2$

$0.2\times0.010 = c_2\times 0.020$

$\implies c_2 = 0.1\mbox{ M}$

Because $\mbox{HCl} \rightarrow \mbox{H}^+ + \mbox{Cl}^-$

$\mbox{[H}^+\mbox{]} = \mbox{[HCl]} = 0.1\mbox{ M}$

$\implies \mbox{pH} = -\log_{10}0.1 = 1$

Question 3

Because $\mbox{NaOH} \rightarrow \mbox{Na}^+ + \mbox{OH}^-$

$\mbox{[OH}^-\mbox{]} = \mbox{[NaOH]} = 10^{-5}\mbox{ M}$

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

$\implies \mbox{[H}^+\mbox{]} = \frac{10^{-14}}{10^{-5}} = 10^{-9}\mbox{ M}$

$\implies \mbox{pH} = -\log_{10}10^{-9} = 9$

Question 4

This is similar to question 2. Try it yourself

My answer was $\mbox{pH} = 0.876$

psychlaw · « **Reply #2 on:** October 09, 2008, 10:48:26 pm »

thanks coblin!

Can someone solve this and just give me the answer to check if I got it right

Q: 50.0 mL of 2M NaOH and 50.0 mL of 0.40M Ca(OH)2
Just the answer

vce08 · « **Reply #3 on:** October 09, 2008, 10:57:35 pm »

What's the question?

psychlaw · « **Reply #4 on:** October 09, 2008, 11:00:38 pm »

ahh shit, I forgot to put the quesiton in

calculate the pH of the following solution:50.0 mL of 2M NaOH and 50.0 mL of 0.40M Ca(OH)2

sorry my bad =/

Collin Li · « **Reply #5 on:** October 09, 2008, 11:01:36 pm »

Since:

$\mbox{NaOH} \rightarrow \mbox{Na}^+ + \mbox{OH}^-$ , and $\mbox{Ca(OH)}_2 \rightarrow \mbox{Ca}^{2+} + 2\mbox{OH}^-$

$\mbox{n(OH}^-\mbox{)} = 0.050\times2 + 2\times0.050\times0.40 = 0.14 \mbox{ mol}$

$\mbox{[OH}^-\mbox{]} = \frac{0.14\mbox{ mol}}{0.050+0.050\mbox{ L}} = 1.4 \mbox{ M}$

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

$\implies \mbox{[H}^+\mbox{]} = \frac{10^{-14}}{1.4} = 7.14\times10^{-15} \mbox{ M}$

$\implies \mbox{pH} = -\log_{10}\left(7.14\times10^{-15}\right) = 14.1$

Pandemonium · « **Reply #6 on:** October 11, 2008, 03:24:23 pm »

Quote from: psychlaw on October 08, 2008, 11:52:31 pm

if you can solve these examples I can do the rest myself (there pretty basic questions)

oh i shouldn't have lolled.

bucket · « **Reply #7 on:** October 11, 2008, 04:51:14 pm »

Quote from: Pandemonium on October 11, 2008, 03:24:23 pm

Quote from: psychlaw on October 08, 2008, 11:52:31 pm
if you can solve these examples I can do the rest myself (there pretty basic questions)
oh i shouldn't have lolled.

lol.................oh god i'm a nerd

Glockmeister · « **Reply #8 on:** October 11, 2008, 07:36:37 pm »

Quote from: bucket on October 11, 2008, 04:51:14 pm

Quote from: Pandemonium on October 11, 2008, 03:24:23 pm
Quote from: psychlaw on October 08, 2008, 11:52:31 pm
if you can solve these examples I can do the rest myself (there pretty basic questions)
oh i shouldn't have lolled.
lol.................oh god i'm a nerd

question is, is the pun intended?

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