Acid-Base Equilib: Solving Homework Problem

AI Thread Summary
To calculate the pH and pOH of a solution created by dissolving 26g of sodium hydroxide in 150mL of water, the correct approach involves determining the number of moles of NaOH and then calculating the molarity. The calculation shows that 0.65 moles of NaOH results in a concentration of approximately 4.33 mol/L, but this value should be adjusted slightly due to the volume change upon dissolution. The pOH is calculated using the concentration of OH- ions, and the pH can then be derived from the relationship pH + pOH = 14. The final pH value calculated is around 13.81, confirming the method is correct when considering the volume of the solution.
Veronica_Oles
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Homework Statement


To unclog a drain you add 26g of sodium hydroxide to 150mL of water. Calculate the pH and pOH for the solution you prepared. Can someone tell me if I'm doing it right.

Homework Equations

The Attempt at a Solution


n = m/M
n = 26/40
n = 0.65mol for 0.15L

Does this mean that I have to convert into mol/L?

So that would be 4.33 mol/L and this would also be my H+ value/ And then would I plug it into my equation

pH = -log[H+] and find my pH value.

Then plug my pH value into my pH + pOH = 14 equation and solve for pOH?

Is this the correct method?
 
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Veronica_Oles said:
So that would be 4.33 mol/L and this would also be my H+ value

No.

Trivial mistake, but you got it reversed. NaOH doesn't produce H+ when dissociating. Other than that, you are on the right track.

Actually the final concentration is a bit lower than 4.33 M as the volume is not 150 mL, it gets a bit higher after NaOH dissolution. But you can safely ignore it.
 
Borek said:
No.

Trivial mistake, but you got it reversed. NaOH doesn't produce H+ when dissociating. Other than that, you are on the right track.

Actually the final concentration is a bit lower than 4.33 M as the volume is not 150 mL, it gets a bit higher after NaOH dissolution. But you can safely ignore it.
Sorry I meant to put OH-

okay here's what I've gotten

NaOH = 26/40 = 0.65mol

And 0.65 mol would be my 0.65 mol/L?
 
Veronica_Oles said:
Sorry I meant to put OH-

okay here's what I've gotten

NaOH = 26/40 = 0.65mol

And 0.65 mol would be my 0.65 mol/L?

I put that value into my pOH equation and ended up with 0.187

I then put that into the next equation
pH = 14 - pOH and got
pH = 13.81

Is this correct or way off?
 
Veronica_Oles said:
NaOH = 26/40 = 0.65mol

And 0.65 mol would be my 0.65 mol/L?

No, you have to take the volume into account, Actually you did it - and correctly - in your first post.
 
Borek said:
No, you have to take the volume into account, Actually you did it - and correctly - in your first post.
Okay so basically I'm using the formula c= n/v?
 
Yes, just like you did initially.
 

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