pH calculation question (1 Viewer)

[Carl5]

This is from the 2008 HSC Paper
Q14, multichoice

20mL of 0.08 mol L-1 HCl is mixed with 30mL of 0.05 mol L-1 NaOH.

What is the pH of the resultant solution

So I found the n of moles for HCl and NaOH using n = c*v. I then found the different. H+ is in excess.
I'm now such at:
c(excess HCl) = 10^-4/volume

How do I get the volume?

 

[UnrealAnchovies]

HCl + NaOH ----> H2O + NaCl

n(HCl) = 0.0016
n(NaOH) = 0.0015
n(excess HCl) = 0.0001
[HCl] = 0.0001 mols / 0.05 L = 0.002 mol/L
pH = -log(0.002) = 2.7 (1 d.p)

 

[Ally14]

I worked out which was in excess ... i.e HCL by 0.0001 moles. As it is a strong monoprotic acid then there will be 0.0001 moles of hydronium ions 50.0 ml of solution. That will give concentration of hydronium ions with c=n/v and then ph=-log[H] . Hope this makes sense , its a bit of work for one multiple choice question ! Hope its not to bad on Wednesday, good luck !

 

[UnrealAnchovies]

Volume is the addition of 30mLs and 20mLs because that's the solution's volume IN TOTAL

 

[Carl5]

Ahhh! That makes sense. Thank you!!