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?

Answer 1

To find the concentration of the 𝐵𝑎(𝑂𝐻)2 solution, we need to use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction:

2 𝐻𝐶𝑂𝑂𝐻 + 𝐵𝑎(𝑂𝐻)2 → 2 𝐻2𝑂 + 𝐵𝑎𝐶𝑂𝑂𝐻

From the balanced equation, we can see that 2 moles of 𝐻𝐶𝑂𝑂𝐻 react with 1 mole of 𝐵𝑎(𝑂𝐻)2.

Let's start by calculating the amount of 𝐻𝐶𝑂𝑂𝐻 reacted in the titration:

Amount of 𝐻𝐶𝑂𝑂𝐻 = concentration of 𝐻𝐶𝑂𝑂𝐻 × volume of 𝐻𝐶𝑂𝑂𝐻
= 0.883 M × 20.4 mL
= 18.0052 mmol

Since 𝐻𝐶𝑂𝑂𝐻 and 𝐵𝑎(𝑂𝐻)2 react in a 1:2 mole ratio, the amount of 𝐵𝑎(𝑂𝐻)2 can be calculated using stoichiometry:

Amount of 𝐵𝑎(𝑂𝐻)2 = 0.5 × amount of 𝐻𝐶𝑂𝑂𝐻
= 0.5 × 18.0052 mmol
= 9.0026 mmol

Finally, we can determine the concentration of the 𝐵𝑎(𝑂𝐻)2 solution as follows:

Concentration of 𝐵𝑎(𝑂𝐻)2 = Amount of 𝐵𝑎(𝑂𝐻)2 / volume of 𝐵𝑎(𝑂𝐻)2
= 9.0026 mmol / 19.3 mL
= 0.466 M

Therefore, the concentration of the 𝐵𝑎(𝑂𝐻)2 solution is 0.466 M.

Answer 2

To find the concentration of the 𝐵𝑎(𝑂𝐻)2 solution, we can use the concept of stoichiometry. The balanced equation for the reaction between 𝑯𝑪𝑶𝑶𝑯 and 𝐵𝑎(𝑂𝐻)2 is:

2𝐻𝐶𝑂𝑂𝐻 + 𝐵𝑎(𝑂𝐻)2 → 𝐵𝑎𝐶𝑂𝑂𝐻2 + 2𝐻2𝑂

From the balanced equation, we can see that the ratio between 𝑜𝑛𝑒 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑯𝐶𝑂𝑂𝐻 and 𝑜𝑛𝑒 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐵𝑎(𝑂𝐻)2 is 2:1.

Given that 20.4 mL of 0.883 M 𝐻𝐶𝑂𝑂𝐻 neutralizes 19.3 mL of 𝐵𝑎(𝑂𝐻)2, we can set up the following equation using the molarity (M) and volume (V) relationship:

M1 x V1 = M2 x V2

Where:
M1 = concentration of 𝑯𝐶𝑂𝑂𝐻 (0.883 M)
V1 = volume of 𝑯𝐶𝑂𝑂𝐻 (20.4 mL)
M2 = concentration of 𝐵𝑎(𝑂𝐻)2 (unknown)
V2 = volume of 𝐵𝑎(𝑂𝐻)2 (19.3 mL)

Plugging in the values we know, we get:

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

Now we can solve for M2:

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

M2 = 0.933 M

Therefore, the concentration of the 𝐵𝑎(𝑂𝐻)2 solution is approximately 0.933 M.

Answer 3

2HCOOH + Ba(OH)2 ==> Ba(OOCH)2 + 2H2O

mols HCOOH = M x L = ?
Looking at the balanced equation,mols Ba(OH)2 is 1/2 mols HCOOH.\
Then M Ba(OH)2 = mols BaOH)2/L Ba(OH)2 = ?