Converting N to Moles in Acid-Base Chemistry

[lha08]

Homework Statement

So I'm trying to solve a problem and it asks: If 1 ml of 10 N NaOH is added to a liter of the buffer prepared, how much will the pH change?
I know how to solve the problem but one thing is really confusing me, how am I able to convert the 1 ml of 10 N NaOH into moles...like what exactly does 'N' signify...? like supposedly it means avagadro's number. The answer to the number of moles is...0.01 mol OH

Homework Equations

The Attempt at a Solution

 

[Borek]

N stands for normality.

 

[lha08]

N stands for normality.

Like it says that normality is equivalent/L...but like how am i supposed to get the NaOH in moles?...Do i calculate it as (10 equivalent/L)*(0.01 L)= 0.1 equivalent...but like what does this mean...?

 

[Borek]

Have you heard about google? Wikipedia?

http://en.wikipedia.org/wiki/Equivalent_(chemistry )

Note that wikipedia uses a narrow definition, that is - equivalents are what reacts with 1 mole of H⁺ or electrons. I prefer other definition - one that takes into account stoichiometry of the given reaction. Hence 1M HCl is always 1N, regardless of the reaction, while 1M H[su]2[/sub]SO₄ is sometimes 1N and sometimes 2N, depending on the reaction.

 

[DeeNos]

To convert 1 ml of 10 N NaOH into moles, we need to understand what "N" signifies. In this context, "N" stands for normality, which is a unit of concentration commonly used in acid-base chemistry. Normality is defined as the number of equivalents of a solute per liter of solution. In this case, the 10 N NaOH solution contains 10 equivalents of NaOH per liter of solution.

To convert from normality to moles, we need to use the following equation:
moles = normality * liters of solution

In this problem, we have 1 ml of 10 N NaOH, which is equivalent to 0.001 liters of solution. Plugging this into the equation, we get:
moles = 10 N * 0.001 L = 0.01 mol OH

So, the answer to the number of moles of OH in 1 ml of 10 N NaOH is 0.01 mol. This means that if 1 ml of 10 N NaOH is added to a liter of the buffer, it will contribute 0.01 mol of OH to the solution, which will affect the pH according to the Henderson-Hasselbalch equation. I hope this helps with your problem!