Question #d327a

2 Answers
Mar 4, 2017

Answer:

There are 0.0039 mol of #"HF"# in the solution.

Explanation:

Step 1. Calculate #["H"_3"O"^"+"]#.

#"pH = 2.70"#

#["H"_3"O"^"+"] = 10^"-pH" color(white)(l)"mol/L" = 10^"-2.70" color(white)(l)"mol/L" = "0.002 00 mol/L"#

Step 2. Calculate the initial value of #"[HF]"#.

We don’t know the initial value of #"[HF]"#.

Let's call it #x# and 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)xcolor(white)(mmmmmmm)0color(white)(mmmmml)0#
#"C/mol·L"^"-1":color(white)(ml)"-0.002 00"color(white)(mmll)"+0.002 00"color(white)(m)"+0.002 00"#
#"E/mol·L"^"-1":color(white)(m) x"-0.002 00"color(white)(mmml)"0.002 00"color(white)(mll)"0.002 00"#

In this case, we know the equilibrium value of #["H"_3"O"^"+"]#, so we work backwards and insert it into the ICE table.

The equilibrium constant expression is

#K_text(a)= (["H"_3"O"^"+"]["F"^"-"])/(["HF"]) = 6.8 × 10^"-4"#

#("0.002 00" × "0.002 00")/(x-"0.002 00") = 6.8 × 10^"-4"#

#4.00 × 10^"-6" = 6.8 × 10^"-4"(x-"0.002 00") = 6.8 × 10^"-4"x - 1.36 ×10^"-6"#

#6.8 × 10^"-4"x = 4.00 × 10^"-6" + 1.36 ×10^"-6"= 5.36 × 10^"-6"#

#x = (5.36 × 10^"-6")/(6.8 × 10^"-4") = 7.88 × 10^"-3"#

#["HF"]_0 = xcolor(white)(l) "mol/L" = 7.88 × 10^"-3"color(white)(l) "mol/L"#

Step 3. Calculate the initial moles of #"HF"#.

#"moles of HF" = 0.500 color(red)(cancel(color(black)("L"))) × (7.88 × 10^"-3"color(white)(l) "mol")/(1 color(red)(cancel(color(black)("L")))) = 3.9 × 10^"-3"color(white)(l) "mol"#

Mar 11, 2017

Here's another way which is related to maths and not chemistry

When we solve for pH we solve like this

Please notice in the calculation I changed pH to 2.63
This is because when I used all the variables including #7.88*10^-3# I found that the pH was 2.63.

#sqrt(Ka*"M") = b#

#"pH" = -log(b)#

Here

#2.636 = "-log"b#

Solve for b
Let's solve your equation step-by-step.
#−log(b)=2.636#
: Divide both sides by -1.

#"−log(b)"/"−1" = (2.636)/(−1)#

#log(b)=−2.635449#
: Solve Logarithm.

#log(b)=−2.635449#

#log(b)=−2.635449#

10log(b)=10−2.636(Take exponent of both sides)

b=#10^(−2.635449#

b=0.002315

#b=0.002315#

Now #K_a = 6.8*10^-4#

#sqrt6.810^−4x=0.002315#

#sqrt0.00068x=0.002315#

Solve Square Root.

#sqrt0.00068x=0.002315#

#0.00068x=0.0023152"(Square both sides)"#

#0.00068x = 0.000005#

#(0.00068x)/0.00068 = 0.000005/0.00068#

(Divide both sides by 0.00068)

x=0.007881

Now I think you know what to do