Question

What is the pH if 1mL of 0.1M HCl is added to 99mL of pure water?

Now if instead of pure water a buffer is used: HPO4-2/H2PO4- pKa = 7.2 Assume the initial pH of this buffer is 7 (like the pure water example). -First you must you must use the Henderson-Hasselbalch equation to determine the ratio of HPO4-2/H2PO4- , which is 0.063M to .1M. Using the same amount of HCl added (.001M), determine the change in pH that occurs to the buffer when the HCl is added.

(i already answered the first part, I just need the second part. Show and explain your work please!)

Answer 1

pH of buffer after addition of 1 mL of 0,1 M HCl = 7,0

Step-by-step explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸; pka=7,2

Thus, Henderson–Hasselbalch equation for 7,00 phosphate buffer is:

7,0 = 7,2 + log₁₀
`([HPO4^(2-)] )/([H2PO4^(-)])`

Ratio obtained is:

0,63 =
`([HPO4^(2-)] )/([H2PO4^(-)])`

As the problem said you can assume [H₂PO₄⁻] = 0,1 M and [HPO4²⁻] = 0,063M

As the amount added of HCl is 0,001 M the concentrations in equilibrium are:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺

0,1 M +x 0,063M -x 0,001M -x -because the addition of H⁺ displaces the equilibrium to the left-

Knowing the equation of equilibrium is:

`K_(a) = ([HPO_(4)^(2-)][H^(+)])/([H_(2) PO_(4)^(-)])`

Replacing:

6,20x10⁻⁸ =
`([0,063-x][0,001-x])/([0,1+x])`

You will obtain:

x² -0,064 x + 6,29938x10⁻⁵ = 0

Thus:

x = 0,063 → No physical sense

x = 0,00099990

Thus, [H⁺] in equilibrium is:

0,001 M - 0,00099990 = 1x10⁻⁷

Thus, pH of buffer after addition of 1 mL of 0,1 M HCl =

-log₁₀ [1x10⁻⁷] = 7,0

A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. In this example you can see its effect!

I hope it helps!