Respuesta :
Answer:
The pH of the solution is 7,22
Explanation:
Using the Henderson–Hasselbalch equation we can find the pH of the solution.
[tex]pH= pKa +log\frac{[A-]}{[HA]}[/tex]
First, we need to find the concentrations of [A-] and [HA]
[tex]7,4=7,2 + log\frac{[A-]}{[HA]} \\\\7,4-7,2 = log\frac{[A-]}{[HA]}[/tex]
Now we apply the antilogarithm function to both sides of the equation
[tex]0,2=log\frac{[A-]}{[HA]} \\\\10^{0,2}=10^{log\frac{[A-]}{[HA]} }\\ \\1,585=\frac{[A-]}{[HA]}[/tex]
using the previous equation and the ratio [A-] + [HA] = 0.1 we proceed to apply a system of equations to find the values of [HA] and [A-].
after solving the system of equations
[A-]=0,0613
[HA]= 0,0387
After the 10 micromol of protons are generated the protons will reduce [A-] and increase [HA].
[H+] = [tex]10^{-6}mol / 1^{-3}L[/tex]
[H+] = 0,01M
now, the new value of pH is
[tex]pH=7,2+log\frac{[0,0613 - 0,01]}{[0,0387 + 0,01]} \\pH=7,22[/tex]
The final pH of the solution has been 7.2.
The pH of the solution has been calculated with the Henderson–Hasselbalch equation as:
[tex]\rm pH=pKa\;+\;log\;\dfrac{[A^-]}{[HA]}[/tex]
Computation for the pH of the solution
The initial pH of the buffer is 7.4
The pKa value of buffer is 7.2
Substituting the values as:
[tex]\rm 7.4=7.2\;+\;log\;\dfrac{[A^-]}{[HA]}\\ log\;\dfrac{[A^-]}{[HA]}=0.2\\\dfrac{[A^-]}{[HA]}=1.585[/tex]
The molarity of the buffer solution has been 0.1 M. Thus,
[tex]\rm [A^-]\;+\;[HA]=0.1[/tex]
Substituting the values for concentration of HA and [tex]\rm A^-[/tex]:
[tex]\rm \dfrac{[A^-]}{1-[A^-]}=1.585\\A^-=0.0613[/tex]
The value of HA has been given as:
[tex]\rm 0.0613\;+\;[HA]=0.1\\HA=0.0387[/tex]
The protons generated in the new reaction has been 10 micromol.
The molarity of the generated protons has been:
[tex]\rm Molarity=\dfrac{Moles}{Volume}\\ Molarity=\dfrac{10^-^6\;mol}{10^-^3\;L}\\ Molarity=0.01\;M[/tex]
The addition of protons results in the decrease in the concentration of [tex]\rm A^-[/tex] and increase in the concentration of HA.
The final pH of the solution has been given as:
[tex]\rm pH=7.2\;+\;log\;\dfrac{0.0613-0.01}{0.0387+0.01}\\ pH=7.2+log\;1\\pH=7.2[/tex]
The final pH of the solution has been 7.2.
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