Solved: Calculate pH of Aqueous Solution w/ HBr(g)

  • Thread starter Thread starter ff_yy
  • Start date Start date
  • Tags Tags
    Aqueous Ph
AI Thread Summary
To calculate the pH of an aqueous solution formed by dissolving HBr gas, one must first determine the moles of HBr present. The original calculation of 37.07 moles is incorrect; under standard conditions, approximately 0.13 moles of HBr can be expected from 3.0L of gas. This results in a concentration of H+ ions that leads to a pH slightly below 0.1. The formula for pH, pH = -log[H+], confirms that negative pH values are possible. Accurate calculations are essential for correct pH determination.
ff_yy
Messages
8
Reaction score
0
[Solved] pH of Aqueous Solution

Homework Statement


Calculate the pH of the following aqueous solution prepared by adding 3.0L HBr(g), measured at 27 degrees Celsius, 757 mmHg pressure, to sufficient water to give 0.10L of solution.


Homework Equations


pH = -log[H+]


The Attempt at a Solution


I'm not sure how to handle a question with gas? Is there another equation I'm supposed to use? The answer says -0.08, which doesn't seem to make sense...

I tried to do this question by first writing the equation
HBr(g) + H20(l) --> Br- + H30+
I found the number of moles of HBr gas to be 37.07...mol, and this is where my understanding fails me and I have no idea what to do. I even tried using n=cv which didn't work...

Please help!
 
Last edited:
Physics news on Phys.org
It's possible to have negative pH. Just look at the formula for pH you have written.

If you have 37.07 mol of HBr, dissolved in water, how much H+ do you have?
 
I'm not going to do the math for you but your answer for moles of HBr is waaaaay too high. Remember that under STP, an ideal gas has but one mole for every 23 L! You are only 2 degrees and 3 mm Hg away from that. Thus, a rough estimate of the number of moles is 3/23 or about 0.13 moles of HBr. In 0.1 liter that would be less than pH 0.1 but not much less.
 
ok thanks, i think i got it...
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top