# Calculate amount of formic acid (ka=2*10-4) which shoud be dissolved in 10L of water to obtain a solution of pH=3.7?

Jun 20, 2018

You should use 0.18 g of formic acid.

#### Explanation:

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

["H"_3"O"^"+"] = 10^"-pH"color(white)(l)"mol/L" = 10^"-3.7"color(white)(l)"mol/L" = 2.0 × 10^"-4"color(white)(l)"mol/L"

Step 2. Calculate the concentration of formic acid

We can use an ICE table to help with the calculation.

$\textcolor{w h i t e}{m m m m m m l} \text{HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m} c \textcolor{w h i t e}{m m m m m m m l} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1": color(white)(ml)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mml)"+} x$
$\text{E/mol·L"^"-1": color(white)(m)c"-} x \textcolor{w h i t e}{m m m m m m m} x \textcolor{w h i t e}{m m m} x$

K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = x^2/(c-x) = 2.0 × 10^"-4"

x^2 = (c-x)(2.0 × 10^"-4") = 2.0 × 10^"-4"c - 2.0 × 10^"-4"x

(2.0 × 10^"-4")c = x^2 + 2.0 × 10^"-4"x

c = (x^2 + 2.0 × 10^"-4"x)/(2.0 ×10^"-4")

x = 2.0 × 10^"-4"

c = ((2.0 × 10^"-4")^color(red)(cancel(color(black)(2))) + (2.0 × 10^"-4")^color(red)(cancel(color(black)(2))))/(color(red)(cancel(color(black)(2.0 × 10^"-4")))) = 4.0 × 10^"-4"

c = 4.0 × 10^"-4"color(white)(l)"mol/L"

Step 3. Calculate the moles of formic acid

$\text{Moles" = 10 color(red)(cancel(color(black)("L"))) × (4.0 × 10^"-4"color(white)(l)"mol")/(1 color(red)(cancel(color(black)("L")))) = 4.0 × 10^"-3"color(white)(l)"mol}$

Step 4. Calculate the mass of formic acid

$\text{Mass" = 4.0 × 10^"-3" color(red)(cancel(color(black)("mol"))) × "46.03 g"/(1 color(red)(cancel(color(black)("mol")))) = "0.18 g}$