# Question d327a

##### 2 Answers
Mar 4, 2017

There are 0.0039 mol of $\text{HF}$ in the solution.

#### Explanation:

Step 1. Calculate $\left[\text{H"_3"O"^"+}\right]$.

$\text{pH = 2.70}$

["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 $\text{[HF]}$.

We don’t know the initial value of $\text{[HF]}$.

Let's call it $x$ and set up an ICE table.

$\textcolor{w h i t e}{m m m m m m m m} \text{HF" + "H"_2"O" ⇌ "H"_3"O"^"+"color(white)(m) + color(white)(ml)"F"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m m l} x \textcolor{w h i t e}{m m m m m m m} 0 \textcolor{w h i t e}{m m m m m l} 0$
$\text{C/mol·L"^"-1":color(white)(ml)"-0.002 00"color(white)(mmll)"+0.002 00"color(white)(m)"+0.002 00}$
$\text{E/mol·L"^"-1":color(white)(m) x"-0.002 00"color(white)(mmml)"0.002 00"color(white)(mll)"0.002 00}$

In this case, we know the equilibrium value of $\left[\text{H"_3"O"^"+}\right]$, 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 $\text{HF}$.

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

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 $7.88 \cdot {10}^{-} 3$ I found that the pH was 2.63.

$\sqrt{K a \cdot \text{M}} = b$

$\text{pH} = - \log \left(b\right)$

Here

$2.636 = \text{-log} b$

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 ${K}_{a} = 6.8 \cdot {10}^{-} 4$

sqrt6.810^−4x=0.002315#

$\sqrt{0.00068} x = 0.002315$

Solve Square Root.

$\sqrt{0.00068} x = 0.002315$

$0.00068 x = 0.0023152 \text{(Square both sides)}$

$0.00068 x = 0.000005$

$\frac{0.00068 x}{0.00068} = \frac{0.000005}{0.00068}$

(Divide both sides by 0.00068)

x=0.007881

Now I think you know what to do