I have troubles with a titration of NaOH and HCl.

The volume of the HCl is 30 ml and its concentration is 0.1 mol per liter
The volume of the NaOH is 25 ml and the concentration is to be worked out. I understand that the reaction ratio for HCl to neutralize NaOH is 1:1.
But does that mean that 0.1 mol of HCl have to react with 0.1 mol of NaOH??? Can someone help me, PLEASE......

Yes, the ratio is 1:1 because of the equation.

HCl + NaOH ==> NaCl + H2O

It means 0.1 mol HCl and 0.1 mol NaOH react exactly; also it means that the moles NaOH must equal the moles of HCl at the equivalence point.
So if you started with 30 mL of 0.1 HCl, you started with M x L = 0.030L x 0.1 M = 0.003 moles HCl. Therefore, at the equivalence point, you must have added 0.003 mole NaOH. Since M = moles/L, 0.003/0.025 (the volume of NaOH) = 0.12 M for the NaOH. For titrations that are 1:1, it is simpler to use mL x M = mL x M
30 x 0.1 = 25 x M and solve for the unknown M.

Thanks!

m1v1=m2v2

Certainly! I can help you with your titration problem.

First, let's understand the reaction that takes place between NaOH and HCl. It is a neutralization reaction, and the balanced equation is:

NaOH + HCl -> NaCl + H2O

According to the balanced equation, the ratio between NaOH and HCl is indeed 1:1. This means that 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl and 1 mole of water.

Now, let's calculate the concentration of the NaOH solution. We can use the concept of mole-to-volume ratio to solve this problem.

Given:
Volume of HCl (VHCl) = 30 mL = 0.03 L
Concentration of HCl (CHCl) = 0.1 mol/L
Volume of NaOH (VNaOH) = 25 mL = 0.025 L

To find the concentration of NaOH (CNaOH), we can use the equation:

CHCl * VHCl = CNaOH * VNaOH

Plugging in the values, we have:

0.1 mol/L * 0.03 L = CNaOH * 0.025 L

Now, we can solve for CNaOH:

CNaOH = (0.1 mol/L * 0.03 L) / 0.025 L

CNaOH = 0.12 mol/L

Therefore, the concentration of the NaOH solution is 0.12 mol/L.

I hope this explanation helps! Let me know if you have any further questions.