pH Calculation (1 Viewer)

micuzzo

Member
Joined
Aug 31, 2008
Messages
489
Gender
Undisclosed
HSC
2009
i need help with this one... caan someone please explain how to do it.

20mL of 0.08mol /L HCl is mixed with 30mL of 0.05mol/L NaOH. What is resultant pH??
 

lolokay

Active Member
Joined
Mar 21, 2008
Messages
1,015
Gender
Undisclosed
HSC
2009
a neutralisation reaction will occur between the NaOH and HCl, so just look at how much of each you have, which is in excess, and what the pH will be due to this excess after neutralisation has completed (I think)
 

gunne

Mew Nember
Joined
Mar 25, 2008
Messages
33
Gender
Male
HSC
2009
Do you mean find molarity or find pH... I think you must mean molarty but i might be wrong
 

micuzzo

Member
Joined
Aug 31, 2008
Messages
489
Gender
Undisclosed
HSC
2009
well u need to find molarity to find the ph... can someone do it
 

GUSSSSSSSSSSSSS

Active Member
Joined
Aug 20, 2008
Messages
1,102
Location
Turra
Gender
Male
HSC
2009
sorry i cbb doin it
but wat u gotta do:

write the reaction between HCl and NaOH

find the limiting reagent

then find the moles of the substance in excess, that is the moles of the limiting reagent subtracted from the moles of the substance in excess

then calculate the concentration of that substance in solution, so dividing it by the volume of the two solutions added together

then u find the concentration of H+ ions/OH- ions, which in this case will be equal to the concentration u just found

and then sub in the formula: -logH+

XDDDD
 

gunne

Mew Nember
Joined
Mar 25, 2008
Messages
33
Gender
Male
HSC
2009
wait you already have molarity of both...
pH = -log(10) (Hydrogen ion concentration)...
does that help
 

undalay

Active Member
Joined
Dec 14, 2006
Messages
1,002
Location
Ashfield
Gender
Male
HSC
2008
20mL of 0.08mol /L HCl is mixed with 30mL of 0.05mol/L NaOH. What is resultant pH??

nHCl = VC = 0.02 x 0.08 = 0.0016 moles
nNaOH = VC = 0.03 x 0.05 = 0.0015

H+ + OH- => H2O (Hydrogen ions and hydroxide ions annihilate.

0.0016-0.0015

remaining = 0.0001 moles of H+
total volume = 0.02+0.03 = 0.05

concentration of H+ remaining = n/v = 0.0001/0.05

-log [H+] = pH

pH = 2.7
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top