chemistry help..standardization of a sodium hydroxide solution?

Question

a total of 23.8 mL of 0.104 M NaOH is used to reach the stoichiometric point in titrating 120.0 ml of a sulfuric acid solution with an unknown concentration. Calculate the molarity of the H2SO4 solution. Write a balanced equation. remember that sulfuric acid is a diprotic acid.

Answer 1

First, this is not a standardization of sodium hydroxide problem. Standardization involves reacting the NaOH with a known amount of acid (usually potassium hydrogen phthalate) to determine its exact concentration.

This is a titration problem
The balanced equation is as follows:
2NaOH + H2SO4 --> 2H2O + Na2SO4

Use the molarity formula to determine how many moles of NaOH were used.
M = moles of NaOH/L of solution

Then use the moles of NaOH with the mole ratio (2:1) from the balanced equation to determine how many moles of H2SO4 reacted.

Divide the number of moles of H2SO4 that reacted by the volume of the H2SO4 in liters to determine the molarity of the acid.

Answer 2

1. Calculate the number of moles of (OH-) from given volume and molarity of NaOH
2. Use mole-mole ratio to calculate molarity of acid (2 to 1 ratio since sulfuric acid has two dissociable protons)

Hope that helps!

Answer 3

A pH of 1 is 10 times as concentrated as a pH of 2 (in terms of acidity). Proof: Use the rule that: pH = -log[H3O+] 1 = -log[H3O+] -1 = log[H3O+] Taking the inverse of log: [H3O+] = 10^-1 M (or 0.1 M) Doing it with a pH of 2: [H3O+] = 10^-2 M (or 0.01 M) 0.1 = ten times larger than 0.01. Therefore a pH of 1 is 10 times as concentrated as a pH of 2 (in terms of acidity). Using the example as above: pH of 12 means [H3O+] = 10^-12 M (This applies to this case because NaOH produces 1 OH- per NaOH). You want it more dilute. Using common sense tells us that adding water in this case would reduce the pH closer to 7. A pH of 7 means that [H3O+] = 10^-7 M. So we want to get closer to 10^-7. What we know: [H3O+] = 10^-12 M We want to get closer to 10^-7 M. 100 times more dilute. How to get there: Multiply 10^-12 by 100. = 10^-10 M pH = -log[H3O+] -log (10^-10) = 10 The new pH is 10.