Question #a721d

`HF + H_2O = H_3O^+ + F^-`

We can find the concentration of `H^+ or H_3O^+` by three ways

One is by the ICE table (but this is a 5% rule) and the other is square root which is absolutely correct and the other is Ostwald's law of dillution

Let's set up an ICE table.

`color(white)(mmmmmmmm)"HF" + "H"_2"O" ⇌ "H"_3"O"^"+"color(white)(m) + color(white)(ml)"F"^"-"`
`"I/mol.L"^"-1":color(white)(mmml)0.88color(white)(mmmmm)0color(white)(mmmmml)0`
`"C/mol.L"^"-1":color(white)(mmml)"-x"color(white)(mmmmll)"+x"color(white)(mlmmml)"+x"`
`"E/mol.L"^"-1":color(white)(mmm) 0.88"- x"color(white)(mmml)"x"color(white)(mmmmmll)"x"`

Ka = `x^2/(0.88-x)`

You can ignore x

Ka = `x^2/(0.88`

`6.8 xx 10 ^-4 = x^2/(0.88-x)`

Step 1: Multiply both sides by -x+0.88.
`−0.00068x+0.000598=x^2`
`−0.00068x+0.000598−x^2=x^2−x^2`(Subtract `x^2` from both sides)
`−x^2−0.00068x+0.000598=0`

x= `{0.0006799999999999999±sqrt(−0.0006799999999999999)^2−4(−1)(0.0005984)}/{2(−1)}`

x = 0.02412457847582909 This is the`H_3O^+`

Another way I told you is very simple

`sqrt(Ka *M) = H_3O^+`

`sqrt (6.8*10^-4 * 0.88M) = 0.024462`

You can see that -x matters so much

Therefore the concentration of HF

0.88M - x(you remember, we did this in the ice table above) = 0.88M - 0.024462M = 0.855538M

You can see that -x matters so much

Third way is

Ostwald's law of dillution

degree of ionization or `alpha` = `sqrt("Ka"/C)`

`H^+ = alpha * C`

Plug in the variables

`sqrt(00068 /0.88) = 0.02780`

`0.02780 * 0.88 = 0.024462M`

Now to calculate `OH^-`

`2H_2O = H_3O + OH^-`

You know that for any substance in water at 298K the
`H_3O^+M * OH^-)M = 1*10^-14`

`0.024462M * (OH^-) = 1*10^-14 = 4.08797E-13M`

`4.08797E-13 = 4.08797 * 10 ^-13M`

Note there are two ways to solve pH . One way is that

`-log( 0.024462) = pH = 1.61151`

`-log(4.08797 * 10 ^-13) = pOH = 12.38849`

And then you can subtract the pOH from 14

And if you add these you get a perfect 14

Method 1. Using the 5 % approximation

Let's set up an ICE table.

`color(white)(mmmmmmmm)"HF" + "H"_2"O" ⇌ "H"_3"O"^"+"color(white)(m) + color(white)(ml)"F"^"-"`
`"I/mol.L"^"-1":color(white)(mml)0.88color(white)(mmmmmml)0color(white)(mmmmmll)0`
`"C/mol.L"^"-1":color(white)(mmll)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mlmmml)"+"x`
`"E/mol.L"^"-1":color(white)(ml) "0.88 -"color(white)(l) xcolor(white)(mmmmml)xcolor(white)(mmxmmm)x`

`K_text(a) = x^2/(0.88-x) = 6.8 × 10^"-4"`

Test for negligibility

`0.88/(6.8 × 10^"-4") = 1300 > 400`

∴ `x` is less than 5 % of the initial concentration of `["HF"]`.

Since `x ≪ 0.88`, it can be ignored in comparison with 0.88.

Then

`x^2/(0.88) = 6.8 × 10^"-4"`

`x^2 = 0.88 × 6.8 × 10^"-4"= 5.98 × 10^"-4"`

`x = 2.45 × 10^"-2"`

`["H"_3"O"^"+"] = x color(white)(l)"mol/L" = "0.024 mol/L"`

`["F"^"-"] = x color(white)(l)"mol/L" = "0.024 mol/L"`

`["OH"^"-"] = K_text(w)/(["H"_3"O"^"+"]) = (1.00 × 10^"-14")/(2.45 ×10^"-2") color(white)(l)"mol/L" = 4.1 × 10^"-13" color(white)(l)"mol/L"`

Method 2. Solving the quadratic

We had the expression

`K_text(a) = x^2/"0.88-x" = 6.8 × 10^"-4"`

`x^2 = 6.8 × 10^"-4"("0.88 -"x) = 5.98 × 10^"-4" - 6.8 × 10^"-4"x`

`x^2 + 6.8 × 10^"-4"x- 5.98 × 10^"-4" = 0`

`x = 2.41 × 10^"-2"`

`["H"_3"O"^"+"] = x color(white)(l)"mol/L" = "0.024 mol/L"`

`["F"^"-"] = x color(white)(l)"mol/L" = "0.024 mol/L"`

`["OH"^"-"] = K_text(w)/(["H"_3"O"^"+"]) = (1.00 × 10^"-14")/(2.41 × 10^"-2") color(white)(l)"mol/L" = 4.1 × 10^"-13" color(white)(l)"mol/L"`

Conclusion

Within the limitations of the allowed significant figures, the approximate method gives the same result as solving the quadratic.