quick ph question

Calculate the pH of the strong acid

10 gdm^-3 HCL

and pH of 0.0400 moldm^-3 Ba(OH)2

Reply 1

Ba(OH)2 = Ba2+ + 2OH-
Therefore c(OH)= 2*0.0400 = 0.0800M
pOH= -log [OH-]
pH= 14-pOH

c(HCl)= 10g/dm3 / 36.46g/mol = 0.274 mol/L
[H3O+] = 0.274M
pH= -log [H30+]

I hope this is correct and will help you.

Reply 2

papz_007

are u sure thats right because it dosnt seem like it is somehow?

Reply 3

tamala

Well, I think it should be OK. It the case with acid, the end pH= 0.562. This is pH of a strong acid and HCl is strong acid.
In the case of Ba(OH)2 is mole of hydorxide releases two moles of OH-. pOH therefore is 1.096 and pH=12.9.
What seems wrong? I won't say that it's 100% right, but I usualy calculate it like this.

Reply 4

papz_007

oh my bad i made a calculator mistake ...sorry thanks for the help tamala

Reply 5

tamala

No problem...

Reply 6

papz_007

hay woundering if u can help me another one

50cm^3 of 0.150moldm^-3 HCl mixed with 500cm^3 of water
wats the pH?

Reply 7

hungry_hog

papz_007

hay woundering if u can help me another one

50cm^3 of 0.150moldm^-3 HCl mixed with 500cm^3 of water
wats the pH?

work out the moles of HCl

no. of moles = concentration (mol/dm3) * volume (dm3)
= 0.150*(50/1000) = 0.0075 moles

now the total volume is 50 + 500 = 550cm3 = 0.55dm3

so the conc of the diluted solution = moles/volume
= 0.0075/0.55 = 0.013636363M

since HCl is a strong acid, can assume 100% dissociated
∴[H+] = 0.013636363M

Ph = -log[H+] = 1.86530143

another method

your starting solution is 0.15M.
it's volume goes up by (500+50)/50 = 11 times
∴conc goes down by 11 times

conc of diluted solution = 0.15/11 = 0.0136363636M

then use -log[H+] as above

don't have a calculator, so you may wish to check the numbers but that's the principle