Finding pH given molarity and volume

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Discussion Overview

The discussion revolves around calculating the pH of a solution formed by mixing 50 ml of 0.004M H2CrO4 with 50 ml of 0.005M HNO3. Participants explore the roles of these substances as acids, the need for dissociation constants, and the implications of strong versus weak acids in the calculation.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant initially identifies H2CrO4 as the acid and HNO3 as the base, questioning the approach taken.
  • Another participant corrects this by stating that both substances are acids, prompting a discussion about their dissociation.
  • It is noted that HNO3 is a strong acid and dissociates completely, while H2CrO4 is not a strong acid, requiring consideration of its dissociation constant (Ka).
  • A calculation is presented for the concentration of H+ ions from HNO3, leading to a proposed pH value, but this is met with skepticism and corrections from other participants.
  • One participant suggests looking up the Ka for H2CrO4 and setting up an equilibrium equation as a necessary step in the calculation.

Areas of Agreement / Disagreement

There is disagreement regarding the classification of the substances involved and the correct approach to calculating pH. Some participants assert that H2CrO4 requires Ka for accurate calculations, while others challenge the calculations presented without reaching a consensus.

Contextual Notes

Participants express uncertainty about the dissociation of H2CrO4 and the implications of using Ka in the calculations. There are unresolved mathematical steps and differing interpretations of the acids' behavior in solution.

Who May Find This Useful

Students working on acid-base chemistry, particularly those dealing with pH calculations involving strong and weak acids.

KaceCC
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Homework Statement


If 50ml of 0.004M H2CrO4 was combined with 50 ml of 0.005M HNO3 what is the pH of the resulting solution?


Homework Equations


Moles H3O=Moles OH ?
Formula for KA



The Attempt at a Solution


So, here's what I have so far.

I made the H2CrO4 represent the acid, and the HNO3 represent the base. I then subtracted them to get the excess, and then found the molarity of that. But I have no idea if I chose the right things to represent acids and bases, in fact I have no idea if I'm doing the right thing. Somebody please help!
 
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Both of these are acids, there is no base here.

Also, do these substances dissociate completely, or must Ka be taken into account?

p.s. Welcome to Physics Forums :smile:
 
Last edited:
I believe HNO3 is a strong acid- you can assume that dissociates completely. However, the other acid is NOT a strong acid--you will need Ka for that.

BUT. I'm assuming since the Ka is very small, adding the H+ ions from that species will not change the # of H+ you get from HNO3 because of sig figs.

Therefore, .05L x .005M H+ (from the complete dissociation of HNO3) = 0.00025mol H+

next, find the molarity. 100ml total = .1L
.0025 Molar H+
-log .0025 = 2.6 = pH
 
Last edited:
cadillac said:
the Ka is very small

No.

.0025 Molar H+

No.

-log .0025 = 2.6 = pH

No.
 
so sad... haha
 
Hell of a thread here.
 
Look up Ka for H2CrO4, and set up the equilibrium equation. That is the key here.
 

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