HX + H2O <---> H3O+ + X- Through previous calculations I found the concentration

Question

HX + H2O <---> H3O+ + X-

Through previous calculations I found the concentrations of:

[HX]= 3.02E-3

[X-]= 3.02E-3

[H3O+]= 9.9E-1

Ka=[H3O+][X-] / [HX] = [3.02E-3] to the second power / 9.9E-1

Now for the second part 20.0 mL of 1.0 M NaX is added to 10 mL of 1.0 M HX. I know the pH of the solution is 4.9, I need to find the following

Concentration of HX after mixing = ?

Concentration of X- after mixing = ?

And

Equilibrium concentration:

[HX]=

[X-]=

[H3O+]= 1.25E-5

Explanation / Answer

FOLLOW THIS SOLVED EXAMPLE First calculate moles of NaOH and moles of HX. moles NaOH = M NaOH x L NaOH = (0.1194)(0.01512) = 0.001805 moles NaOH. moles HX = M HX x L HX = (0.1246)(0.02500) = 0.003115 moles HX. HX and NaOH react as follows: HX + NaOH ==> H2O + NaX Since the HX and NaOH react in a 1:1 mole ratio, then the 0.003115 moles of HX will completely use up the 0.001805 moles of NaOH. You will have (0.003115 - 0.001805) = 0.001310 moles of HX left and you will have formed 0.001805 moles of NaX. The combination of HX and NaX is a buffer system. The pH of this buffer is given by the Henderson-Hasselbalch equation: pH = pKa + log (moles NaX / moles HX) 5.9 = pKa + log (0.001805 / 0.001310) 5.9 = pKa + 0.139 5.76 = pKa Ka = 10^-5.76 = 1.7 x 10^-6

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