Question

When 0.1M of HCI is reacted with 0.2M of NaOh, it from NaOh . What will be the concentration of NaOh?
Give HCI=2ml, ​

Answer 1

Explanation:

Calculate the concentration of the hydrochloric acid.

Volume of sodium hydroxide solution = 25.00 ÷ 1000 = 0.0250 dm 3

Amount of sodium hydroxide = 0.200 × 0.0250 = 0.005 mol.

From the equation, 0.005 mol of NaOH reacts with 0.005 mol of HCl.

Volume of hydrochloric acid = 22.70 ÷ 1000 = 0.0227 dm 3

Answer 2

Answer:

Explanation:

!! EXTREMELY LONG ANSWER !!

You're titrating hydrochloric acid, HCl, a strong acid, with sodium hydroxide, NaOH, a strong base, so right from the start you should know that the pH at equivalence point must be equal to 7.

Hydrochloric acid and sodium hydroxide react in a 1:1 mole ratio to form water and aqueous sodium chloride

HCl(aq]+NOH(aq]→NaCl(aq]+H2O(l]

The net ionic equation for this reaction looks like this

H3O+(aq]+OH−(aq]→2H2O(l]

Now, the equivalence point corresponds to a complete neutralization, i.e. when you add enough strong base to completely consume all the acid present in the solution.

The 1:1 mole ratio tells you that at the equivalence point, the solution must contain equal numbers of moles of strong acid and strong base.

Your tool of choice here will the equation

∣∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aapH=−log([H3O+])aa∣∣−−−−−−−−−−−−−−−−−−−−−−−−

Now, it's very important to realize that the volume of the solution will increase as you add the strong base solution. Keep this in mind when calculating the molarity of the hydronium ions.

So, let's start calculating the corresponding pH

1.Before NaOH is added−−−−−−−−−−−−−−−−−−−

Since hydrochloric acid is a strong acid, it dissociates completely in aqueous solution to form hydronium cations and chloride anions. More specifically, you have

[H3O+]=[HCl]=0.1 M

This means that the pH of the solution before any strong base is added will be equal to

pH=−log(0.1)=∣∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯aa1aa∣∣−−−−−−