Can someone please answer one of these question so I would be able to know how to do the others?

1. What volume of 0.275 M KOH (aq) must be added to 75.0 mL of 0.137 M H3AsO4 (aq) to reach (a) the first stoichiometric pony; (b) the second stoichiometric point

2. What volume of 0.0848 M HCL must be added to 88.8 mL of 0.233 M Na3PO4 to reach (a) the first stoichiometric point; (b) the second stoichiometric point; (c) the third stoichiometric point

3. What volume of 0.255 M HNO3 must be added to 35.5 mL of 0158 M Na2HPO3 to reach (a) the first stoichiometric point; (b) the second stoichiometric point

4. What volume of 0.123 M NaOH(aq) must be added to 125 mL of 0.197 M H2SO3(aq) to reach (a) the first stoichiometric point; (b) the second stoichiometric point

Thank you so much!

Write and balance the equation.

H3AsO4 + KOH ==> KH2AsO4 + H2O
mols H3AsO4 = M x L = ?
Now use the coefficients in the balanced equation to convert mols H3AsO4 to mols KOH. That's 1:1 so mols H3AsO4 = mols KOH.
Then M KOH = mols KOH/L KOH. You have mols KOH and M KOH, solve for L KOH and convert to mL if that's what you need. Note that this is the volume required to neutralize the first H of H3AsO4. It will take that same amount to neutralize the second H and that same amount again to neutralize the third H.

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To solve these questions, we'll be using the concept of stoichiometry in chemical reactions. Stoichiometry is a calculation of the quantities of reactants and products involved in a chemical reaction.

The first step is to write out the balanced chemical equation for each reaction. The coefficients in the balanced equation will help us determine the stoichiometric ratios between the reactants.

Once we have the balanced equation, we can use the equation:
(moles of solute) = (concentration of solute) * (volume of solution in liters)

We can then use this formula to calculate the moles of each reactant. To find the volume of the solution needed for the reaction, we can rearrange the formula to solve for volume:
(volume of solution in liters) = (moles of solute) / (concentration of solute)

Now let's go through each question step by step:

1. First, we need to write the balanced equation for the reaction:
H3AsO4 + 3KOH -> KH2AsO4 + 2H2O

(a) To reach the first stoichiometric point, we need to add enough KOH to completely react with all the H3AsO4. The moles of H3AsO4 can be calculated using the formula mentioned above:
moles of H3AsO4 = (0.137 M) * (0.0750 L)

Since the stoichiometric ratio between H3AsO4 and KOH is 1:3, we need three times as many moles of KOH. Therefore:
moles of KOH = 3 * moles of H3AsO4

Now, we can calculate the volume of 0.275 M KOH solution needed:
(volume of KOH solution) = (moles of KOH) / (0.275 M)

(b) To reach the second stoichiometric point, we need to add enough KOH to react with all the KH2AsO4 formed in the first reaction. The moles of KH2AsO4 can be calculated using the formula mentioned above:
moles of KH2AsO4 = (0.137 M) * (0.0750 L) * 2

Again, using the stoichiometric ratio between KH2AsO4 and KOH (1:3), we can calculate the moles of KOH required:
moles of KOH = 3 * moles of KH2AsO4

The volume of 0.275 M KOH solution needed can be calculated as:
(volume of KOH solution) = (moles of KOH) / (0.275 M)

2. Following the same approach as above, let's write the balanced equation for the reaction:
3HCl + Na3PO4 -> 3NaCl + H3PO4

(a) To reach the first stoichiometric point, we need to react all the Na3PO4 with HCl. Calculate the moles of Na3PO4 first:
moles of Na3PO4 = (0.233 M) * (0.0888 L)

The stoichiometric ratio between Na3PO4 and HCl is 1:3, so we need three times as many moles of HCl. Therefore:
moles of HCl = 3 * moles of Na3PO4

Now calculate the volume of 0.0848 M HCl solution needed:
(volume of HCl solution) = (moles of HCl) / (0.0848 M)

(b) To reach the second stoichiometric point, we will use the same approach as in part (a), but this time the stoichiometric ratio is 1:1 between NaCl and H3PO4. Calculate the moles of H3PO4:
moles of H3PO4 = (0.233 M) * (0.0888 L)

The volume of 0.0848 M HCl solution needed can be calculated as:
(volume of HCl solution) = (moles of H3PO4) / (0.0848 M)

(c) We can solve part (c) using the same steps as above, but this time the stoichiometric ratio is 3:1 between H3PO4 and HCl.

3. Let's write the balanced equation for the reaction:
HNO3 + Na2HPO3 -> NaH2PO3 + H2O

(a) Calculate the moles of Na2HPO3 using the formula mentioned earlier:
moles of Na2HPO3 = (0.158 M) * (0.0355 L)

The stoichiometric ratio between Na2HPO3 and HNO3 is 1:1, so the moles of HNO3 needed will be the same as the moles of Na2HPO3:
moles of HNO3 = moles of Na2HPO3

Now calculate the volume of 0.255 M HNO3 solution needed:
(volume of HNO3 solution) = (moles of HNO3) / (0.255 M)

(b) The stoichiometric ratio between NaH2PO3 and HNO3 is 1:1, so the moles of HNO3 needed will be the same as the moles of NaH2PO3 in this case. Calculate the volume of 0.255 M HNO3 solution needed using the formula:
(volume of HNO3 solution) = (moles of NaH2PO3) / (0.255 M)

4. The balanced equation for this reaction is:
H2SO3 + 2NaOH -> Na2SO3 + 2H2O

(a) To reach the first stoichiometric point, calculate the moles of H2SO3:
moles of H2SO3 = (0.197 M) * (0.125 L)

The stoichiometric ratio between H2SO3 and NaOH is 1:2, so we need twice as many moles of NaOH. Therefore:
moles of NaOH = 2 * moles of H2SO3

Calculate the volume of 0.123 M NaOH solution needed:
(volume of NaOH solution) = (moles of NaOH) / (0.123 M)

(b) To reach the second stoichiometric point, we use the same steps as in part (a), but this time the stoichiometric ratio is 1:1 between Na2SO3 and H2SO3.

Note: I have provided the general steps and calculations for each question. Make sure to double-check the calculations and units to ensure accuracy.

I hope this explanation helps you understand how to approach these types of questions.