Question

Question #d327a

Answer 1

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"`

Answer 2

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