In a titration experiment, 20.4 mL of 0.883 M 𝐻𝐶𝑂𝑂𝐻 neutralize 19.3 mL of 𝐵𝑎(𝑂𝐻)2. What is the concentration of the 𝐵𝑎(𝑂𝐻)2 solution?

Question

In a titration experiment, 20.4 mL of 0.883 M 𝐻𝐶𝑂𝑂𝐻 neutralize 19.3 mL of 𝐵𝑎(𝑂𝐻)2. What is the concentration of the 𝐵𝑎(𝑂𝐻)2 solution?

Do I use M1V1=M2V2? my answer is 0.467, is this correct?

Answer 1

Here is my work on the problem yesterday.

https://www.jiskha.com/questions/1823891/in-a-titration-experiment-20-4-ml-of-0-883-m-neutralize-19-3-ml-of

Answer 2

No. You do not use M1V1 = M2V2. That formula works very well for titrations that are 1:1; i.e., say NH3 + HCl ==> NH4Cl. It does NOT work for reactions that are not 1:1: i.e., 2HCOOH + Ba(OH)2 ==> Ba(OOCH)2.

By the way I worked this problem for you yesterday. The answer to day is the same. I will try to find it and post it with a link for you.

Answer 3

Ohhh, sorry i already forgot that i already ask the question, i thought no one answer the previous one so i asked it again. thank you dr bob,

Answer 4

Well, at least you're not trying to titrate clown noses! But let's break down this problem, shall we?

In this titration, we're looking at the reaction between 𝐻𝐶𝑂𝑂𝐻 and 𝐵𝑎(𝑂𝐻)2.

Using the equation M1V1=M2V2, where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume, we can solve for M2.

Since we're neutralizing 𝐻𝐶𝑂𝑂𝐻 with 𝐵𝑎(𝑂𝐻)2, we know that the moles of acid will equal the moles of base:

M1V1 = M2V2

(0.883 M) * (20.4 mL) = M2 * (19.3 mL)

Now, let's calculate M2:

M2 = (0.883 M * 20.4 mL) / 19.3 mL

After doing the math, the concentration of the 𝐵𝑎(𝑂𝐻)2 solution is approximately 0.933 M, not 0.467. So, it seems like you've got a double dose of clowning around!

Keep up the good work, and don't forget to laugh along the way!

Answer 5

Yes, you are correct in using the formula M1V1 = M2V2 for this problem. The formula relates the concentration (Molarity) and volume of solutions before and after mixing.

Let's calculate the concentration of the Ba(OH)2 solution:

M1 = concentration of HCOOH = 0.883 M
V1 = volume of HCOOH = 20.4 mL = 0.0204 L
V2 = volume of Ba(OH)2 = 19.3 mL = 0.0193 L
M2 = concentration of Ba(OH)2 (what we want to find)

Using the formula M1V1 = M2V2, we can substitute these values:

(0.883 M)(0.0204 L) = (M2)(0.0193 L)

Now, rearrange the equation to solve for M2:

M2 = (0.883 M)(0.0204 L) / (0.0193 L)
= 0.009322 / 0.0193
≈ 0.482 M

So, the concentration of the Ba(OH)2 solution is approximately 0.482 M, not 0.467 M.