Answer:
The pH of the mixture is approximately 4.2.
There's no equilibrium calculation involved if HCl is used instead of lactic acid. The pH of that solution will be approximately 3.7.
Explanation:
Lactic acid is a monoprotic acid. In other words, each lactic acid molecule can be ionized to release only one proton. Therefore, lactic acid will react with NaOH at a 1-to-1 ratio. The result will be one unit of the anion of lactate ion.
Lactic acid is in excess. All NaOH will be consumed. Before any dissociation takes place, the solution will contain 0.2 mM of lactic acid and 0.4 mM of lactate ions.
Lactic acid is a weak acid. The following equilibrium will take place in the solution:
[tex]\rm C_3H_6O_3 \leftrightharpoons {C_3H_5O_3}^{-} + H^{+}[/tex].
Let the increase in the concentration of protons in the solution be [tex]x\rm \;M[/tex]. Construct a RICE table:
[tex]\begin{array}{c|ccccc}\rm \textbf{R}&\rm C_3H_6O_3 & \leftrightharpoons &\rm {C_3H_5O_3}^{-}& +& \rm H^{+}\\\textbf{I} & 0.2\times 10^{-3} && 0.4\times 10^{-3} \\\textbf{C} & -x & & +x & & +x\\\textbf{E} & 0.2\times 10^{-3} -x & & 0.4\times 10^{-3} + x & & x\end{array}[/tex].
The pKa value of lactic acid is 3.86. Therefore, [tex]K_{\rm a} = 10^{-3.86}[/tex]. By the definition of the acid dissociation constant [tex]K_{\rm a}[/tex],
[tex]\begin{aligned} K_{\rm a} &= \frac{\rm [H^{+}][{C_3H_5O_3}^{-}]}{[\rm C_3H_6O_3]}\\& = \frac{x(0.4\times 10^{-3} +x)}{(0.6 \times 10^{-3} - x)}\end{aligned}[/tex].
Given that [tex]K_{\rm a} = 10^{-3.86}[/tex], solve for [tex]x[/tex]:
[tex]\displaystyle \frac{x(0.4\times 10^{-3} - x)}{0.2 \times 10^{-3}-x} = 10^{-3.86}[/tex].
There might be more than one solutions. However, the correct solution shall ensure that all concentrations are greater than zero. In this case,
[tex]x \approx 5.7\times 10^{-5}[/tex].
By the definition of pH,
[tex]\rm pH = -\log_{10}{[H^{+}]} = -\log_{10}{5.7\times 10^{-5}} \approx 4.2[/tex].
The Henderson-Hasselbalch equation will yield a similar result.
Similar to lactic acid, HCl is also monoprotic. The solution will contain 0.2 mM of HCl and 0.4 mM of NaCl before any ionization takes place.
Unlike lactic acid, all that 0.2 mM of HCl will ionize to produce protons and chloride ions. The final proton concentration in the solution will be 0.2 mM. As a result,
[tex]\rm pH = - \log_{10}{[H^{+}]} = -\log_{10}{0.2\times 10^{-3}} \approx 3.6[/tex].
We have that the pH of a mixture of equal volumes of 0.6 mM Lactic acid and 0.4 mM NaOH is
pH=4.0
From the question we are told that
0.6 mM Lactic acid
0.4 mM NaOH
Generally the equation for the moles of HCl is mathematically given as
moles of HCl=m*v=0.6m*v
Where The Chemical equation is given as
HCl+NaOH->NaCl+H20
Hence Ratio is given as
0.2:0:0.4
Where
NaCl has no effect on pH as a Neutral salt
[tex]HCl=\frac{moles of HCl}{V}\\\\HCl=\frac{0.2}{2}\\\\HCl=0.1mm[/tex]
Therefore
[tex]pH=-log{H^+}\\\\\pH=-log{0.1*10^3}[/tex]
pH=4.0
Therefore
The pH of a mixture of equal volumes of 0.6 mM Lactic acid and 0.4 mM NaOH is
pH=4.0
For more information on this visit
https://brainly.com/question/17756498?referrer=searchResults
