Answer:
pH of buffer after addition of 1 mL of 0,1 M HCl = 7,0
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 ![\frac{[HPO4^{2-}] }{[H2PO4^{-}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHPO4%5E%7B2-%7D%5D%20%7D%7B%5BH2PO4%5E%7B-%7D%5D%7D)
Ratio obtained is:
0,63 =
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 -<em>because the addition of H⁺ displaces the equilibrium to the left-</em>
Knowing the equation of equilibrium is:
![K_{a} = \frac{[HPO_{4}^{2-}][H^{+}]}{[H_{2} PO_{4}^{-}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BHPO_%7B4%7D%5E%7B2-%7D%5D%5BH%5E%7B%2B%7D%5D%7D%7B%5BH_%7B2%7D%20PO_%7B4%7D%5E%7B-%7D%5D%7D)
Replacing:
6,20x10⁻⁸ = ![\frac{[0,063-x][0,001-x]}{[0,1+x]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B0%2C063-x%5D%5B0%2C001-x%5D%7D%7B%5B0%2C1%2Bx%5D%7D)
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!
