Question

1. the solubility of co₂ gas in water is 25°c and 1 atm is 0.034 mol/l. what is the solubility under atmospheric conditions? (the partial pressure of co₂ in air is 0.0003 atm.) assume that co₂ obeys henry’s law. show all work! 6. calculate the molarity of a phosphoric acid (h₃po₄) solution that is 84% by mass phosphoric acid and has a density of 1.87 g/ml.

Explanation:

Step1: Identify Henry's Law formula

Henry's Law is $C = kP$, where $C$ is solubility, $k$ is Henry's - law constant, and $P$ is partial pressure. First, find $k$ using the given data. At $P_1 = 1$ atm and $C_1=0.034$ mol/L, $k=\frac{C_1}{P_1}$.
$k=\frac{0.034\ mol/L}{1\ atm}=0.034\ mol/(L\cdot atm)$

Step2: Calculate solubility at new pressure

We know the partial pressure of $CO_2$ in air $P_2 = 0.0003$ atm. Using Henry's Law $C_2=kP_2$.
$C_2 = 0.034\ mol/(L\cdot atm)\times0.0003\ atm=1.02\times 10^{-5}\ mol/L$

for question 6:

Step1: Assume a volume of solution

Let's assume we have $V = 1\ L=1000\ mL$ of the solution.

Step2: Calculate the mass of the solution

Given the density $
ho = 1.87\ g/mL$, the mass of the $1 - L$ solution $m=
ho V$.
$m = 1.87\ g/mL\times1000\ mL = 1870\ g$

Step3: Calculate the mass of phosphoric acid

The solution is 84% by mass phosphoric acid. So the mass of $H_3PO_4$, $m_{H_3PO_4}=0.84\times1870\ g = 1570.8\ g$

Step4: Calculate the molar mass of phosphoric acid

The molar mass of $H_3PO_4$: $M=(3\times1 + 31+4\times16)\ g/mol=98\ g/mol$

Step5: Calculate the number of moles of phosphoric acid

$n=\frac{m_{H_3PO_4}}{M}=\frac{1570.8\ g}{98\ g/mol}=16.03\ mol$

Step6: Calculate the molarity

Molarity $M=\frac{n}{V}$, and since $V = 1\ L$, the molarity of the $H_3PO_4$ solution is $16.03\ mol/L$

Answer:

$1.02\times 10^{-5}\ mol/L$