[n4mel355]

Hi
My supervisor asked me to demonstrate for some second year students at uni. Problem is I have never been super strong at maths and it has been over half a decade since I even used this stuff. I can do most of the stuff but one question that the students have been assigned I cant seem to get my head around.

Q
Howwould you make 100mL of 50mM Tris/HCl? pH 7.5 (MWt of Tris base is 121.1 g/mol), and has a pKa of 8.3 at RT. The concentration of HCL is 11.5M

I can easily enough figure out the amount of Tris but get stuck figuring out the amount of HCL required. The answer is as follows

pH-pKa+log([base]/[acid])
7.5=8.3+log([base]/[acid])
log([base]/[acid])= -0.8
[base]/[acid]= 0.1585  (I can get up to here with no problem, and understand that this is the ratio of base to acid)

[base]+[acid]= 50mM
(0.1585+1)[acid]= 50mM  

(This is where I am unsure, I understand that the 1 is from the ratio of 0.1585/1 and that you assume that acid and base is equal to 1 + 0.1585 .

So does [acid]+[base]=1.1585 and as [acid]+[base]=50mM does (1/1.1585)50mM=acid and (0.1585/1.1585)50mM=[base]

(1/1.1585)50mM=43.16mM (??molarity of H30+??)
(0.1585/1.1585)50mM= 6.84mM

From here on its simple dilution that I understand.
Could someone please shed some light and better explain this in a way that I could easily explain it to the students.

Cheers

[UG]

[base]/[acid]= 0.1585             [1]
[base]+[acid]= 50mM              [2]
Rearrange [1] to get [base] = 0.1585[acid]
Substitute into [2] to get: 0.1585[acid] + [acid] = 50 mM
Factorise to get: [acid](1+0.1585) = 50 mM

[n4mel355]

OMG taking out the common factor, wow learnt that back in middle school.

Cheers for this.