Finding pH given molarity and volume

AI Thread Summary
To find the pH of a solution created by mixing 50 ml of 0.004M H2CrO4 and 50 ml of 0.005M HNO3, it is essential to recognize that both substances are acids, with HNO3 being a strong acid that dissociates completely. The calculation involves determining the total moles of H+ ions contributed by HNO3, which is 0.00025 mol, leading to a concentration of 0.0025 M H+. However, since H2CrO4 is a weak acid, its dissociation constant (Ka) must be considered to accurately calculate the pH. The final pH is derived from the concentration of H+ ions, resulting in a pH of approximately 2.6. Understanding the role of both acids and their dissociation is crucial for solving this problem.
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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
 
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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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