The concentration of an unknown acid solution with general formula H3A is to be determined by titration with a 0.850 M KOH solut
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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
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:
Replacing:
6,20x10⁻⁸ =
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!
Answer:
Explanation:
From the given information:
At wavelength = 270 nm
At 270 nm
Suppose x is said to be the solution for the concentration of x and y to be the solution for the concentration of y;
Then:
Divide both sides by 200
Use l = 1cm (i.e the standard length)
Then;
<u>For 540 nm:</u>
since l = 1
Equating both (1) and (2) together, we have: