Question

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

Answer 1

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 $\frac{[HPO4^{2-}] }{[H2PO4^{-}]}$

Ratio obtained is:

0,63 = $\frac{[HPO4^{2-}] }{[H2PO4^{-}]}$

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:

$K_{a} = \frac{[HPO_{4}^{2-}][H^{+}]}{[H_{2} PO_{4}^{-}]}$

Replacing:

6,20x10⁻⁸ = $\frac{[0,063-x][0,001-x]}{[0,1+x]}$

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!