I need help on these three questions. For all three of them you're supposed to write a balanced equation for each neutralization. Then, to find the concentrations of the unknown solutions if the following reactions are found to reach equivalence by titration.

The three questions are:
5) 21.7 mL of 0.0500 M Ba(OH)2 is added to 35.0 mL of HCl of unknown concentration.
6) 25.0 mL of 0.200 M lithium hydroxide is added to 20.0 mL of phosphoric acid of unknown concentration.
7) 30.0 mL of 0.500 M hydrobromic acid is added to 38.0 mL of strontium hydroxide of unknown concentration.

First, there is no way for me to know if these titrations have reached the equivalence point since the concn is unknown; however, I suspect what you really want to know is the concn of the unknown material

Ba(OH)2 + 2HCl ==> 2H2O + BaCl2
mols Ba(OH)2 = M x L = 0.0217 x 0.05 = apprx 0.00108
Using the coefficients in the balanced equation, convert mols Ba(OH)2 to mols HCl. That's 0.00108 mols Ba(OH)2 x (2 mols HCl/1 mol Ba(OH)2) = 0.00216
Then M HCl = mols HCl/L HCl = 0.00216/0.035 = apprx 0.062 but you need to go through and get these numbers more accurately.
6 and 7 are done the same way but I would point out on #6 that you don't indicate how many of the H ions in H3PO4 have been titrated.

Sure, here are the balanced equations for each neutralization reaction:

5) Ba(OH)2 + 2HCl -> BaCl2 + 2H2O

6) 3LiOH + H3PO4 -> Li3PO4 + 3H2O

7) 2HBr + Sr(OH)2 -> SrBr2 + 2H2O

To find the concentrations of the unknown solutions, we need to use the concept of stoichiometry.

For question 5, we can use the balanced equation to determine the ratio between the moles of Ba(OH)2 and HCl. From the equation, we can see that 1 mole of Ba(OH)2 reacts with 2 moles of HCl. By using the volume and the concentration of Ba(OH)2, we can calculate the number of moles of Ba(OH)2. Then, using the stoichiometric ratio, we can determine the number of moles of HCl. Finally, by dividing the moles of HCl by the volume of HCl used, we can find the concentration of the unknown HCl solution.

The same approach can be applied to questions 6 and 7 to find the concentrations of the unknown solutions. Just follow the steps mentioned earlier for each question using the given information of volumes and concentrations.

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

Sure! Let's solve each question step-by-step.

5) The balanced equation for the neutralization reaction between Ba(OH)2 and HCl is:

Ba(OH)2 + 2HCl -> BaCl2 + 2H2O

To find the concentration of HCl, we can use the equation:

M1V1 = M2V2

Where M1 is the concentration of Ba(OH)2, V1 is the volume of Ba(OH)2, M2 is the concentration of HCl, and V2 is the volume of HCl.

Given:
M1 = 0.0500 M
V1 = 21.7 mL
V2 = 35.0 mL

Plugging in the values into the equation, we get:

(0.0500 M)(21.7 mL) = (M2)(35.0 mL)

Simplifying, we have:

M2 = (0.0500 M)(21.7 mL) / (35.0 mL)
M2 = 0.0311 M

Therefore, the concentration of HCl is approximately 0.0311 M.

6) The balanced equation for the neutralization reaction between lithium hydroxide (LiOH) and phosphoric acid (H3PO4) is:

3LiOH + H3PO4 -> Li3PO4 + 3H2O

To find the concentration of H3PO4, we can use the equation:

M1V1 = M2V2

Given:
M1 = 0.200 M
V1 = 25.0 mL
V2 = 20.0 mL

Plugging in the values into the equation, we get:

(0.200 M)(25.0 mL) = (M2)(20.0 mL)

Simplifying, we have:

M2 = (0.200 M)(25.0 mL) / (20.0 mL)
M2 = 0.250 M

Therefore, the concentration of H3PO4 is approximately 0.250 M.

7) The balanced equation for the neutralization reaction between hydrobromic acid (HBr) and strontium hydroxide (Sr(OH)2) is:

2HBr + Sr(OH)2 -> SrBr2 + 2H2O

To find the concentration of Sr(OH)2, we can use the equation:

M1V1 = M2V2

Given:
M1 = 0.500 M
V1 = 30.0 mL
V2 = 38.0 mL

Plugging in the values into the equation, we get:

(0.500 M)(30.0 mL) = (M2)(38.0 mL)

Simplifying, we have:

M2 = (0.500 M)(30.0 mL) / (38.0 mL)
M2 = 0.395 M

Therefore, the concentration of Sr(OH)2 is approximately 0.395 M.

To answer these questions, we need to first write balanced equations for each neutralization reaction.

For question 5:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O

For question 6:
2LiOH + H3PO4 -> Li3PO4 + 3H2O

For question 7:
2HBr + Sr(OH)2 -> SrBr2 + 2H2O

Now, let's use the balanced equations to find the concentrations of the unknown solutions using the concept of titration and equivalence points.

For question 5:
Given:
Volume of Ba(OH)2 solution (V1) = 21.7 mL
Concentration of Ba(OH)2 solution (C1) = 0.0500 M
Volume of HCl solution (V2) = 35.0 mL
Unknown concentration of HCl solution (C2)

The balanced equation gives us the stoichiometric ratio between Ba(OH)2 and HCl as 1:2. So, the moles of Ba(OH)2 and HCl are equal at equivalence point.

Moles of Ba(OH)2 = C1 * V1
Moles of HCl = 2 * Moles of Ba(OH)2

Now, we can set up an equation using the equation moles = concentration * volume:

C1 * V1 = 2 * C2 * V2

Substituting the given values, we get:

0.0500 M * 21.7 mL = 2 * C2 * 35.0 mL

Solving for C2, we find:

C2 = (0.0500 M * 21.7 mL) / (2 * 35.0 mL)

C2 ≈ 0.0179 M

So, the concentration of the unknown HCl solution is approximately 0.0179 M.

Now, let's move to question 6:

Given:
Volume of LiOH solution (V1) = 25.0 mL
Concentration of LiOH solution (C1) = 0.200 M
Volume of phosphoric acid solution (V2) = 20.0 mL
Unknown concentration of phosphoric acid solution (C2)

The balanced equation gives us the stoichiometric ratio between LiOH and H3PO4 as 2:1. So, the moles of LiOH and H3PO4 are equal at equivalence point.

Moles of LiOH = 2 * Moles of H3PO4

Using the equation moles = concentration * volume, we can set up the following equation:

2 * C1 * V1 = C2 * V2

Substituting the given values, we have:

2 * 0.200 M * 25.0 mL = C2 * 20.0 mL

Solving for C2, we find:

C2 = (2 * 0.200 M * 25.0 mL) / 20.0 mL

C2 = 0.500 M

So, the concentration of the unknown phosphoric acid solution is 0.500 M.

Finally, let's tackle question 7:

Given:
Volume of HBr solution (V1) = 30.0 mL
Concentration of HBr solution (C1) = 0.500 M
Volume of Sr(OH)2 solution (V2) = 38.0 mL
Unknown concentration of Sr(OH)2 solution (C2)

The balanced equation gives us the stoichiometric ratio between HBr and Sr(OH)2 as 2:1. So, the moles of HBr and Sr(OH)2 are equal at equivalence point.

Moles of HBr = Moles of Sr(OH)2 / 2

Using the equation moles = concentration * volume, we can set up the following equation:

C1 * V1 = C2 * V2 / 2

Substituting the given values, we get:

0.500 M * 30.0 mL = C2 * 38.0 mL / 2

Solving for C2, we find:

C2 = (0.500 M * 30.0 mL * 2) / 38.0 mL

C2 ≈ 0.789 M

So, the concentration of the unknown Sr(OH)2 solution is approximately 0.789 M.