# Solving a pH Change After Adding Sulfuric Acid to Sodium Hydroxide

• arctic_viper
In summary: I'll try to remember that in the future.In summary, In a 50.0 mL aqueous solution of sodium hydroxide, adding 36.00 ml of 0.0200 mol/L sulfuric acid will cause the pH to increase to 12.51.
arctic_viper

## Homework Statement

A 50.0 mL aqueous solution of sodium hydroxide has a pH of 12.50. If 36.00 ml of 0.0200 mol/L sulfuric acid is added to this sodium hydroxide solution, what will be the new pH of the resulting solution? Assume that the temperature stays constant at 25C, and that the volumes are perfectly additive.

2. The attempt at a solution
I tried looking at a similar post -->https://www.physicsforums.com/threads/quick-ph-calculation-please-verify.753708/#post-5639696, (I have no idea how he/she did it)

I've tried doing 10^-12.50 to get [OH-] but I'm not sure it's right.

3. I have no idea what to do to solve this question. Please help, and thank you. :D

What you did was more or less a correct first step - that should give you the concentration of NaOH.

Now write the reaction equation.

So,
2NaOH(aq) + H2SO4(aq) ---> Na2SO4(aq) +2H2O(l) ?

How many moles of the NaOH were present? What is the limiting reagent?

Please try to apply what you know to the problem, if you are going to just do whatever you are told you will never learn how to move ahead.

arctic_viper said:
I've tried doing 10^-12.50 to get [OH-] but I'm not sure it's right.

Will there is a way of being more sure one way or the other. In a neutral pure water [OH-] is 10-7. So for an NaOH solution does 10-12.5 sound reasonable?

It would be nice and helpful also to alingy him/herself if he/she came back and explained

arctic_viper
Are you familiar with the fact that $$log[H^+]+log[OH^-]=-14$$ and that $$pH=-log[H^+]$$

arctic_viper
epenguin said:
Will there is a way of being more sure one way or the other. In a neutral pure water [OH-] is 10-7. So for an NaOH solution does 10-12.5 sound reasonable

It would be nice and helpful also to alingy him/herself if he/she came back and explained

Yes, sorry, I had my exam yesterday and was really tired. But I did find the concentration by doing the following.

14-12.50 =1.50

then 10^-1.50 to get 3.16x10^-2

Then I divided that by the 50.00ml of water to find initial moles.

After doing that I made an IRF table..I also found the moles of H2SO4 by using the given information. I pretty much just solved the IRF table. (I'd have to look at my notes again to see what else I did.) But I think I undertand how to solve the problem. Thanks and sorry.

arctic_viper said:
Yes, sorry, I had my exam yesterday and was really tired. But I did find the concentration by doing the following.

14-12.50 =1.50

then 10^-1.50 to get 3.16x10^-2

Then I divided that by the 50.00ml of water to find initial moles.
You should multiply by 0.05 liters of water to find the initial moles. The 0.0316 has units of moles/liter.

Chestermiller said:
You should multiply by 0.05 liters of water to find the initial moles. The 0.0316 has units of moles/liter.
Ya that's pretty much what I did. I converted ml to liters by dividing by 1000. After that I multiplied the [OH-] by the L to get the moles of [OH-]

Checking for reasonableness and orders of magnitude needs to become a second-nature habit for chemical calculations. Even if you were partly along the right lines humourless Profs etc. may give you no credit at all or even less if your answer is out by 10 orders of magnitude!

epenguin said:
Checking for reasonableness and orders of magnitude needs to become a second-nature habit for chemical calculations. Even if you were partly along the right lines humourless Profs etc. may give you no credit at all or even less if your answer is out by 10 orders of magnitude!
What do you mean by orders of magnitude?

arctic_viper said:
What do you mean by orders of magnitude?

Factors of 10

epenguin said:
Factors of 10
I see. Thanks.

## 1. How does the addition of sulfuric acid affect the pH of sodium hydroxide?

When sulfuric acid is added to sodium hydroxide, a neutralization reaction occurs. This results in the formation of water and a salt, which can impact the pH of the solution. The exact change in pH will depend on the concentrations of the acid and base, as well as the volume of each added.

## 2. What is the equation for the neutralization reaction between sulfuric acid and sodium hydroxide?

The neutralization reaction between sulfuric acid and sodium hydroxide can be represented by the following equation: H2SO4 + 2NaOH → 2H2O + Na2SO4. This equation shows the formation of water and sodium sulfate, a salt, from the reactants sulfuric acid and sodium hydroxide.

## 3. How can the change in pH be calculated after adding sulfuric acid to sodium hydroxide?

To calculate the change in pH after adding sulfuric acid to sodium hydroxide, you will need to know the initial concentrations of the acid and base, as well as the volume of each added. Then, you can use the equation for the neutralization reaction to determine the moles of each reactant and product. Finally, you can use the Henderson-Hasselbalch equation or a pH calculator to determine the change in pH.

## 4. Will the pH change be the same if the order of addition is reversed?

No, the order of addition can impact the resulting pH change. When sulfuric acid is added to sodium hydroxide, the neutralization reaction will occur quickly and the pH will change. However, if sodium hydroxide is added to sulfuric acid, the reaction will take longer and the pH change may not be as significant.

## 5. How can the pH be adjusted back to its original level after adding sulfuric acid to sodium hydroxide?

To adjust the pH back to its original level, you can add a weak acid or base to the solution, depending on the direction of the pH change. For example, if the pH decreases after adding sulfuric acid, you can add a weak base like sodium bicarbonate to increase the pH. Alternatively, you can dilute the solution by adding more water to decrease the concentration of the acid or base.

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