2. 12.5 ml of 0.280M HNO3 and 5.0mL of 0.920M KOH are mixed. To make the resulting solution neutral, which of the following should be added?

-1.1 mol HCl
-1.69 mL of 0.650M HBr
-0.55 mmol Mg(OH)2
-2.2mL of 0.50M LiOH
-none of the above; already neutral

-KOH is in excess by 0.0011mol but if I divide by the molar mass of KOH I get 1.96*10^-5g. None of the answers are right, where do I take it from here?

The excess is 0.001mol but this answer is in L so if you multiply it by 1000 to make your answer in mol/mL like the original question, you will get the answer 1.1mol/mL and that is the first choice: 1.1mol HCl.

To determine which compound needs to be added to neutralize the solution, we need to consider the chemical equation between HNO3 and KOH. The balanced equation is:

HNO3 + KOH → KNO3 + H2O

From the equation, we can see that 1 mole of HNO3 reacts with 1 mole of KOH to form 1 mole of KNO3 and 1 mole of water.

Given that we have 12.5 mL of 0.280 M HNO3, we can calculate the number of moles of HNO3:

(0.280 mol/L) * (12.5 mL / 1000 mL) = 0.0035 mol HNO3

Given that we have 5.0 mL of 0.920 M KOH, we can calculate the number of moles of KOH:

(0.920 mol/L) * (5.0 mL / 1000 mL) = 0.0046 mol KOH

Since the stoichiometry of the equation is 1:1 between HNO3 and KOH, and the number of moles of KOH is greater than the number of moles of HNO3, it means that the KOH is in excess and not all of it will react.

Now, if none of the options provided are correct, we need to find an alternative solution. One possible way to make the resulting solution neutral is by adding a weak base that can react with the excess acid (HNO3). One option would be to add an equivalent amount of a weak base like sodium bicarbonate (NaHCO3).

The balanced equation between NaHCO3 and HNO3 is:

NaHCO3 + HNO3 → NaNO3 + H2O + CO2

By adding NaHCO3, it will react with the excess HNO3 to form NaNO3, water, and carbon dioxide. This will neutralize the acidic solution.

Therefore, the correct answer would be none of the provided options, and instead, sodium bicarbonate (NaHCO3) should be added to make the resulting solution neutral.

To determine the correct answer, we need to consider the balanced chemical equation for the reaction between HNO3 and KOH:

HNO3 + KOH → KNO3 + H2O

From the equation, we can see that 1 mole of HNO3 reacts with 1 mole of KOH to produce 1 mole of water.

Given that 12.5 mL of 0.280 M HNO3 is used, we can calculate the number of moles of HNO3:

moles of HNO3 = volume (L) x concentration (mol/L)
= 0.0125 L x 0.280 mol/L
= 0.0035 mol HNO3

Similarly, for 5.0 mL of 0.920 M KOH:

moles of KOH = volume (L) x concentration (mol/L)
= 0.005 L x 0.920 mol/L
= 0.0046 mol KOH

Now, we need to determine the limiting reactant, which is the one that will be completely consumed in the reaction. The stoichiometry of the balanced equation tells us that 1 mole of HNO3 reacts with 1 mole of KOH.

Comparing the number of moles of HNO3 and KOH, we can see that there is an excess of KOH by:

excess KOH = moles of KOH - moles of HNO3
= 0.0046 mol KOH - 0.0035 mol HNO3
= 0.0011 mol KOH

Since we are looking for a substance to add to the mixture, we need to consider only the options with the least amount of substance to react with the excess KOH, which is 0.0011 mol KOH.

Now, let's consider the answer choices:

-1.1 mol HCl: Since 1 mole of HCl is equal to 1 mole of KOH, this option is not feasible because it exceeds the amount needed (0.0011 mol).

-1.69 mL of 0.650 M HBr: To determine the number of moles of HBr, we can use the concentration and volume:

moles of HBr = volume (L) x concentration (mol/L)
= 0.00169 L x 0.650 mol/L
= 0.0010975 mol HBr

Comparing the number of moles of HBr with the excess KOH, we find that this option is also not feasible because it exceeds the amount needed (0.0011 mol).

-0.55 mmol Mg(OH)2: To convert mmol to mol, we divide by 1000:

moles of Mg(OH)2 = 0.55 mmol / 1000
= 0.00055 mol Mg(OH)2

Comparing the number of moles of Mg(OH)2 with the excess KOH, we find that this option is also not feasible because it exceeds the amount needed (0.0011 mol).

-2.2 mL of 0.50 M LiOH: To determine the number of moles of LiOH, we can use the concentration and volume:

moles of LiOH = volume (L) x concentration (mol/L)
= 0.0022 L x 0.50 mol/L
= 0.0011 mol LiOH

Comparing the number of moles of LiOH with the excess KOH, we find that this option matches the amount needed (0.0011 mol).

Therefore, 2.2 mL of 0.50 M LiOH should be added to the mixture to make the resulting solution neutral.

It's worth noting that this calculation assumes ideal conditions and doesn't take into account any possible side reactions or changes in volume upon mixing.