Acid-Base Equilib: Solving Homework Problem

In summary, to calculate the pH and pOH of a solution made by adding 26g of sodium hydroxide to 150mL of water, you first need to calculate the number of moles of NaOH using the formula n=m/M. This results in 0.65 mol of NaOH for 0.15L. Then, using the equation pH = -log[H+], you can calculate the pH value by converting the concentration to mol/L. However, since NaOH does not produce H+ when dissociating, you need to use the OH- concentration instead. Finally, to find the pOH, you can use the equation pOH + pH = 14 and solve for pOH. It is important
  • #1
Veronica_Oles
142
3

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?
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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?
 
  • #4
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?
 
  • #5
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.
 
  • #6
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?
 
  • #7
Yes, just like you did initially.
 

FAQ: Acid-Base Equilib: Solving Homework Problem

What is an acid-base equilibrium?

An acid-base equilibrium is a chemical reaction between an acid and a base that results in the formation of water and a salt. This reaction involves the transfer of protons (H+) from the acid to the base, leading to the formation of a conjugate acid-base pair.

How do you determine the pH of a solution in an acid-base equilibrium?

The pH of a solution in an acid-base equilibrium can be determined by using the Henderson-Hasselbalch equation, which takes into account the concentration of the acid, the concentration of the base, and the dissociation constant of the acid. Alternatively, you can use a pH meter to directly measure the pH of the solution.

What is a titration and how is it used to solve acid-base equilibrium problems?

A titration is a laboratory technique used to determine the concentration of an unknown solution by reacting it with a known solution of a strong acid or base. In acid-base equilibrium problems, a titration can be used to determine the concentration of the acid or base present in a solution, or to find the equilibrium constant of the reaction.

How do you calculate the equilibrium constant for an acid-base equilibrium?

The equilibrium constant for an acid-base equilibrium can be calculated using the concentrations of the reactants and products at equilibrium. For a general acid-base reaction, the equilibrium constant, also known as the acid dissociation constant (Ka), is equal to the concentration of the products divided by the concentration of the reactants.

What are some common mistakes to avoid when solving acid-base equilibrium problems?

Some common mistakes to avoid when solving acid-base equilibrium problems include not correctly identifying the acid and base in the reaction, not using the correct equilibrium constant, and not considering the effect of dilution on the concentrations of the reactants and products. It is also important to double-check your calculations and units to ensure accuracy.

Similar threads

Chemistry Moles of NaOH to Create Buffer Solution pH=6 in 0.42M Ethanoic Acid
Replies
3
Views
2K
Calculating pH and pOH for 0.048 M Benzoic Acid Solution
Replies
4
Views
3K
Titration of Sulfurous Acid: Calculating pH with Ka Values
Replies
3
Views
2K
How Do You Calculate the pH at Equivalence Points in Malonic Acid Titration?
Replies
6
Views
3K
Titration of amino acid with strong base
Replies
16
Views
4K
Solving a pH Change After Adding Sulfuric Acid to Sodium Hydroxide
Replies
12
Views
4K
Titration of acetic acid w/ NaOH problem
Replies
2
Views
2K
I am sure I am doing this Acid/Base problem wrong
Replies
3
Views
2K
How Do You Calculate pH and [OH-] for a Mixed NH3 and NH4Cl Solution?
Replies
1
Views
1K
Acids and Bases calculation problems
Replies
4
Views
5K

Hot Threads

Recent Insights

Back
Top