A solution is made by mixing 12.0 g of NaOH and 75.0 ml of 0.200 M HNO3

a) write a balanced equation for the reaction that occurs between the solutes.
b) Caclulate the concentration of each ion remaining in solution.
c) Is the resultant solution acidic or basic?

O.k. I am having a hard time with this problem.
for a) I wrote:

NaOH + HNO3 ---> NaNO3 + H20

The rest I have no clue and that's were I need help!

a- NaOH + HNO3 ---> NaNo3+ H20

b- Naoh= 0.300 Mole
Hno3= .015

c- basic

Oh, I see you're having a "solute-ion" relationship problem. Don't worry, I'll help you solve it with a touch of clown humor!

a) Your balanced equation is almost there! It should be:

2NaOH + HNO3 ---> Na2NO3 + 2H2O

b) Now, let's calculate the concentration of each ion remaining in solution. Since we have 75.0 mL of 0.200 M HNO3, we can use the equation to find the moles of HNO3:

0.0750 L x 0.200 mol/L = 0.015 mol HNO3

Since there is 2:1 stoichiometry between HNO3 and NaOH, we will need half the amount of NaOH:

0.015 mol HNO3 / 2 = 0.0075 mol NaOH

Now that we know the moles of NaOH, we can calculate the concentration of each ion remaining:

For Na+: 0.0075 mol NaOH / 0.0750 L = 0.100 M Na+

For NO3-: 0.015 mol HNO3 / 0.0750 L = 0.200 M NO3-

c) Is the resultant solution acidic or basic? Well, since we started with a strong base (NaOH) and a weak acid (HNO3), the resulting solution will be slightly basic. But don't worry, it won't turn your tongue blue or anything!

Great start on part (a)! Your balanced equation is correct:

NaOH + HNO3 ---> NaNO3 + H2O

Now let's move on to part (b) to determine the concentration of each ion remaining in solution. To do this, we need to consider the stoichiometry of the reaction and the given information.

First, let's find the moles of NaOH and HNO3:

Moles of NaOH = Mass of NaOH / Molar mass of NaOH
Moles of NaOH = 12.0 g / 40.00 g/mol = 0.300 mol

Moles of HNO3 = Volume of HNO3 x Concentration of HNO3
Moles of HNO3 = 75.0 mL x 0.200 mol/L = 15.0 mmol = 0.0150 mol

From the balanced equation, we can see that 1 mole of HNO3 reacts with 1 mole of NaOH to produce 1 mole of NaNO3. So, all of the NaOH will react with HNO3, and 0.300 mol of NaNO3 will be formed.

Now let's find the concentration of Na+ ions and NO3- ions in the resultant solution:

Concentration of Na+ ions = Moles of NaNO3 / Volume of solution
Concentration of Na+ ions = 0.300 mol / 75.0 mL = 4.00 M

Concentration of NO3- ions = Moles of NaNO3 / Volume of solution
Concentration of NO3- ions = 0.300 mol / 75.0 mL = 4.00 M

So, the concentration of each ion remaining in solution is 4.00 M.

Finally, let's move to part (c) and determine whether the resultant solution is acidic or basic. Since there are no excess H+ ions or OH- ions present, the solution will be neutral. NaNO3 is a neutral salt and does not significantly contribute to the acidity or basicity of the solution.

Therefore, the resultant solution is neutral.

I hope this explanation helps! If you have any further questions, feel free to ask.

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