question from past paper (1 Viewer)

[teurino]

how do u go about answering this question coz im just revising for half-yearly atm . If anyone can show me the procedure of solving the question it will be much appreciated. Its from HSC past paper 07'
Question 10 MC

10​

[FONT=LBDLH J+ Times,Times]A 0.1 mol L[/FONT]<SUP>[FONT=LBDLH J+ Times,Times]–1 [/FONT][FONT=LBDLH J+ Times,Times]HCl solution has a pH of 1.0 .​

<DIR><DIR>What volume of water must be added to 90 mL of this solution to obtain a final pH of 2.0?
</DIR></DIR>

1. (A) 10 mL
2. (B) 180 mL
3. (C) 810 mL
4. (D) 900 mL

</SUP>[/FONT]

 

[minijumbuk]

C)

pH=-log [H+]

10^-pH = [H+]
[H+] = 10^-2
[H+] = 0.01 M, or 0.01 mol/L

This means that the final solution should have a concentration of 0.01 M

In 90 mL of 0.1 M HCl, the number of moles of H+ ions = 0.09 x 0.1
Which equals 0.009 moles

Now that you have the number of moles of HCl you have, and the concentration you have to make it to, you can work out the volume.

c = n/v
v = n/c
v = 0.009 moles / 0.01 mol/L
v = 0.9 L

Therefore, FINAL volume of solution is 900 mL. However, do not forget that you already have 90 mL before the dilution.

So the amount of water you add is 900 - 90

Which gives the answer 810 mL, which is C.

 

[undalay]

super quick method:

pH 1 = 90 mL of solution
pH 2 = ??

pH 2 = 100x less concentration as pH 1
so pH2 solution must be 10x as dilute

so volume must be 10x as much.

as 900-90 = 810 mL must be added.

 

[minijumbuk]

super quick method:

pH 1 = 90 mL of solution
pH 2 = ??

pH 2 = 100x less concentration as pH 1
so pH2 solution must be 10x as dilute

so volume must be 10x as much.

as 900-90 = 810 mL must be added.

Brilliant, as always =]