Calculating pH

[luken93]

(Soz to hijack your thread)

Also, I got a couple of questions.

If it's a diprotic such as H2SO4 do you double the [H+] concentration? Can't remember if I'm thinking of the wrong thing...

Also, I had a question today that I can't remember how to do.

250ml of .0500M H2SO4
a) find pH
b) If 1L of water is added,what is the new pH
c) Find the Volume of NaOH needed to equalise the solution?

Yeah, you double the concentration.

a) pH=-log(0.1)=1
b)  Find mol H+
     n(H+)=cv=0.1x0.25=0.025 mol
     Find new conc of H+
     c=n/v=0.0025/1.25=0.02
     pH=-log(0.02)=1.70
c)  H2SO4 + 2NaOH  ---> Na2SO4 + 2H2O
     Just realised you can't do it without the concentration of the NaOH...

Thread hijack? Don't be silly. There was a largeish question on pH/concentrations with a diprotic acid. Finding this out last night I thinks given me a real boost Even though it's not important, year 11 and all, I still like to do reasonably well.

haha no worries, happy to help haha

[ruchika5]

Hey all, I have a few questions related to calculating the pH of a solution.

1) What is the pH of a solution produced with 3.5g of nitric acid dissolves in 150mL water?

2) What is the pH of a solution when 70mL of 0.2M HCl is added to 30mL 0.5M barium hydroxide solution?

3)What mass of HCl must be added to 80mL 0.5M potassium hydroxide to lower the pH to 5?

Thanks!

[schnappy]

1)
HNO3

n(H+)=n(HNO3)=3.5/63=0.0555556
[H+]=n/v=0.0555556/0.150=0.3703704
pH=-log(0.3703704)=profit

[thushan]

1)

So you want to first know the amount of HNO3 per litre. --> need n(HNO3)

n(HNO3) = m/M = 3.5/(1.0 + 14.0 + 16.0 x 3) = 0.056 mol
[H+] = [HNO3] = 0.056/0.150 = 0.37 M (n = cV)
=> pH = -log[10] (0.37) = 0.431

2)

need [H+] or [OH-]

n(OH-)[after reaction] = 0.030 x 0.5 x 2 - 0.070 x 0.2 = 0.016 mol
[OH-] = 0.016/0.100 = 0.16 M
=> pH = 13.2

3)

you want [H+] = 10^(-5) => n(H+) = 8 x 10^(-7)

The odd thing about this question is that the above number, representing the excess [H+], is very very very small. This is basically equivalent to adding equilvalent amount of HCl for neutralisation.

in that case you'd get 1.46 g of HCl. If you really want to be precise to get a pH of 5, you need 1.4600292 g.  

[schnappy]

Wow now I feel lazy :/

[nacho]

you're both lazy bums for not using latex

[thushan]

We're all lazy!

[Mao]

you're both lazy bums for not using latex

[pi]

Alternatively
-log[OH]
= x

pH=14 - x

Simplify to pH = 14 + log[OH].

Saves a step