Whilst doing buffer capacity, my lecturer presented us with the following equation for buffer capacity:

$${log(\frac{[salt] + x}{[acid] - x})} = 1$$

A task set by the lecturer is to rearrange the above equation to make $x$ the subject, when $\ce[salt] = [acid]$. I am struggling with this a little, and after searching the internet for help, I can't seem to come across the equation at hand.

Here's what I tried:

  • Took antilog of both sides: $\frac{[salt] + x}{[acid] - x} = 10$
  • Cross-multiplied: $[salt] + x = 10.[acid] - x$
  • Collected like-terms: $[salt] + 2x = 10.[acid]$
  • Solved for $x$: $x = 5.([acid] - [salt])$

However, the above is incorrect, and apparently it should be: $x = \frac{10.([acid] - [salt])}{11}$ which I simply have no idea to get to. Bear in mind my mathematics isn't that great, unfortunately.

I also attempted to include a little reasoning based on the Henderson-Hasselbalch equation.

I assumed that when $[salt] = [acid]$, $log(\frac{[salt]}{[acid]}) = 0$ in the Henderson-Hasselbalch equation, and so maybe $log(\frac{[salt]}{[acid]}) + 1 = {log(\frac{[salt] + x}{[acid] - x})}$. However, this gets me nowhere either.

Does anyone have any thoughts on this, as to how I could arrive at the correct expression for $x$? Thanks.

  • 1
    $\begingroup$ My dear, You have done the 2nd step of cross multiplication wrong. You have written it as, $[salt] + x = 10.[acid] -x$. But it should be $[salt] + x = 10.[acid] - 10x$. Then do the rest calculation, you will get your desired answer. $\endgroup$ – Soumik Das Jan 16 at 12:30
  • $\begingroup$ Solving for x would tell you what concentration of strong base would have to be in the solution in order to set the pH at one unit above the pKa of the buffer, no matter what the current concentrations and pH. Not sure how this is related to buffer capacity, though. $\endgroup$ – Karsten Theis Jan 16 at 15:03
  • $\begingroup$ If you do the algebra carefully, your result should be $x = \frac{10[acid] - [base]}{11}$, i.e. [base] is not multipled by 10. $\endgroup$ – Karsten Theis Jan 16 at 15:05
  • $\begingroup$ @Karsten Theis I've arrived at that expression now, thanks. I suppose it relates to buffer capacity enough though. x in this case is the buffer capacity. My lab work today was to investigate this actually - turns out x is the theoretical buffer capacity. $\endgroup$ – undergraduate839283 Jan 16 at 19:39

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