Question #a721d

2 Answers
Mar 17, 2017

Answer:

pH = 1.61151
#OH^- = 4.08797 * 10 ^-13M#
HF = 0.855538M
#H^+ = 0.024462M#
#F^- = 0.024462M#

Explanation:

#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

Mar 17, 2017

Answer:

#["H"_3"O"^"+"] = "0.024 mol/L"#
#"[F]"^"-"color(white)(mm) = "0.024 mol/L"#
#["OH"^"-"] color(white)(m)= 4.1 × 10^"-13" color(white)(l)"mol/L"#

Explanation:

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.