Question

Due to a highway accident, 150 L of concentrated hydrochloric acid (12.0 M) is released into a lake containing 5.0 ´ 105 m3 of water. If the pH of this lake was 7.0 prior to the accident, what is the pH of the lake following the accident?

Answer 1

Answer: The pH of the lake after accident is 5.44

Explanation:

To calculate the molarity of lake, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the water

$M_2\text{ and }V_2$ are the molarity and volume of hydrochloric acid

We are given:

Conversion factor used:  $1m^3=1000L$

$M_1=?M\\V_1=(5\times 10^5m^3+150L)=[(5\times 10^8)+(150)]L\\M_2=12.0M\\V_2=150L$

Putting values in above equation, we get:

$M_1\times(5\times 10^8+150)=12.0\times 150\\\\M_1=\frac{12.0\times 150}{(5\times 10^8+150)}=3.59\times 10^{-6}$

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

$[H^+]=3.59\times 10^{-6}$

Putting values in above equation, we get:

$pH=-\log(3.59\times 10^{-6})\\\\pH=5.44$

Hence, the pH of the lake after accident is 5.44

Answer 2

Answer:

The pH of the lake following the accident is 5.43.

Explanation:

$Molarity=\frac{Moles}{Voluem(L)}$

Concentration of HCL =12.0 M

Volume of HCL = 150 L

Moles of hydrogen ions= n

$n=12 M\times 150 L=1800 mol$

The pH of the lake = 7

$pH=-\log[H^]$

$7=-\log[H^+]$

$[H^+]=10^{-7} M$

Concentration of hydrogen ions in lake = $10^{-7} M$

Volume of water in lake  = $V_2=5\times 10^5 m^3 = 5\times 10^8 L$

Moles of hydrogen ions in lake = n'

$n'=10^{-7} M\times 5\times 10^8 L=50 mol$

Concentration of hydrogen ions after release of HCL in lake: C

Total moles of HCl = n + n' = 1800 mol + 50 mol = 1850 mol

Volume of the lake = $150 L + 5\times 10^8 L$

$C=\frac{1850 mol}{150 L + 5\times 10^8 L}=3.6999\times 10^{-6} M$

The final pH of the lake:

$pH=-\log[C]$

=$-\log[3.6999\times 10^{-6} M]=5.43$

The pH of the lake following the accident is 5.43.