Question

Calculate the equilibrium concentrations of Na+, Cl−,H+, CHO−2, and HCHO2 when 50.0 mL of 0.25 M HCl is mixed with 50.0 mL of 0.25 M NaCHO2 (?)

Answer 1

You have not quoted `pK_a` for formic acid.

Explanation:

In aqueous solution formate ion undergoes the equilibrium:

`HCO_2^(-) + H_2O(l) rightleftharpoons HCO_2H(aq) + HO^-`

However, as described this solution is stoichiometric in formic acid. The solution will be acidic depending on the magnitude of `K_a`....and so....

`HCO_2H(aq) + H_2O(l) rightleftharpoons HCO_2^(-)+ H_3O^+`

And given `pK_a=3.75`...then `K_a=10^(-3.75)=1.78xx10^-4`... And so in the usual way...

`([HCO_2^(-)][H_3O^+])/([HCO_2H])=1.78xx10^-4`...and if the amount of base that associates is `x`...then....

`(x^2)/(0.125-x)=1.78xx10^-4`...and if `0.125">>"x`...then...

`x=sqrt(1.78xx10^-4xx(0.125-x))`...and making approximations...

`x_1=4.71xx10^-3*mol*L^-1`...

`x_2=4.62xx10^-3*mol*L^-1`...

`x_3=4.64xx10^-3*mol*L^-1`...

`x_4=4.63xx10^-3*mol*L^-1`...

`x_5=4.63xx10^-3`...the values have converged, and I am prepared to accept this as the true value. And so `x=[H_3O^+]=4.63xx10^-3*mol*L^-1`..and you can fill in the rest of the required values....

Answer 2

Warning! Long Answer. Here's what I get.

Explanation:

The equilibrium concentrations are:

`["Na"^"+"] color(white)(ml)= "0.125 mol/L"`
`["Cl"^"-"]color(white)(mml)= "0.125 mol/L"`
`["H"^"+"]color(white)(mm) = 4.7 × 10^"-3"color(white)(l)"mol/L"`
`["HCHO"_2^"-"] = 4.7 × 10^"-3"color(white)(l)"mol/L"`
`["HCHO"_2] = "0.120 mol/L"`

Initial moles of each species

In 50.0 mL of 0.25 mol/L `"HCl"`,

`"Amount of HCl" = 50.0 color(red)(cancel(color(black)("mL HCl"))) × ("0.25 mmol HCl")/(1 color(red)(cancel(color(black)("mL HCl")))) = "12.5 mmol HCl"`

#"Amount of Cl"^"-" = 12.5 color(red)(cancel(color(black)("mmol HCl"))) × ("1 mmoL H"^"+") /(1 color(red)(cancel(color(black)("mmol HCl"))))= "12.5 mmol Cl"^"-"#

The equilibrium calculation

In 50.0 mL of 0.25 mol/L `"NaCHO"_2`,

`"Amount of NaCHO"_2 = 50.0 color(red)(cancel(color(black)("mL NaCHO"_2))) × ("0.25 mmol NaCHO"_2)/(1 color(red)(cancel(color(black)("mL NaCHO"_2)))) = "12.5 mmol NaCHO"_2`

`"Amount of Na"^"+" = 12.5 color(red)(cancel(color(black)("mmol NaCHO"_2))) × (1 "mmol Na"^"+")/(1 color(red)(cancel(color(black)("mmol NaCHO"_2)))) = "12.5 mmol Na"^"+"`

`color(white)(mmmmmm)"HCl" + "NaCHO"_2 → "HCHO"_2 + "H"_2"O"`
`"I/mmol": color(white)(mll)12.5 color(white)(mmll)"12.5color(white)(mmmmm)0`
`"C/mmol": color(white)(m)"-12.5"color(white)(mml)"-12.5"color(white)(mmml)"+12.5"`
`"E/mmol": color(white)(mm)0color(white)(mmmml)0color(white)(mmmmm)12.5`

Equilibrium concentrations

We have a solution that contains `"12.5 mmol HCHO"_2` in 100.0 mL of solution.

`["HCHO"_2] = "12.5 mmol"/"100.0 mL" = "0.125 mmol/mL = 0.125 mol/L"`

The equilibrium involved is

`"HCHO"_2 +"H"_2"O" ⇌ "H"_3"O"^"+" + "CHO"_2^"-"; K_text(a) = 1.77 × 10^"-4"`

We can use another ICE table to calculate the concentrations of the species involved.

`color(white)(mmmmmml)"HCHO"_2 +"H"_2"O" ⇌ "H"_3"O"^"+" + "CHO"_2^"-"`
`"I/mol·L"^"-1": color(white)(mll)0.125 color(white)(mmmmmmml)0color(white)(mmmm)0`
`"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmmmmml)"+"xcolor(white)(mmm)"+"x`
`"E/mol·L"^"-1": color(white)(m)"0.125-"xcolor(white)(mmmmmml)xcolor(white)(mmmm)x`

`K_text(a) = (["H"_3"O"^+]["HCHO"_2^"-"])/(["HCHO"_2]) = x^2/(0.125-x) = 1.77 × 10^"-4"`

Check for negligibility

`0.125/(1.77 × 10^"-4") = 706 > 400`. ∴ `x ≪ 0.125`.

Then,

`x^2/0.125 = 1.77 × 10^"-4"`

`x^2 = 0.125 × 1.77 × 10^"-4" = 2.21 × 10^"-5"`

`x = 4.7 × 10^"-3"`

Calculate the concentration of each species

`["Na"^"+"] = "12.5 mmol"/"100 mL"= "0.125 mol/L"`

`["Cl"^"-"] = "12.5 mmol"/"100 mL"= "0.125 mol/L"`

`["H"^"+"] = xcolor(white)(l)"mol/L" = 4.7 × 10^"-3"color(white)(l)"mol/L"`

`["HCHO"_2^"-"] = xcolor(white)(l)"mol/L" = 4.7 × 10^"-3"color(white)(l)"mol/L"`

`"HCHO"_2] = (0.125 - x) color(white)(l)"mol/L" = (0.125 - 4.7 × 10^"-3")color(white)(l) "mol/L = 0.120 mol/L"`