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!

calculate the moles of each reactant. The smaller of the two is how much reacted.

a) The balanced equation for the reaction that occurs between NaOH and HNO3 is:

NaOH + HNO3 -> NaNO3 + H2O

b) To calculate the concentration of each ion remaining in solution, we need to determine the number of moles of NaOH and HNO3 initially present in the solution, and then use stoichiometry to determine the number of moles of Na+ and NO3- ions.

Step 1: Calculate the number of moles of NaOH:
molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.00 g/mol
moles of NaOH = mass / molar mass = 12.0 g / 39.00 g/mol = 0.308 mol

Step 2: Calculate the number of moles of HNO3:
moles of HNO3 = volume (in liters) * molarity
volume of HNO3 = 75.0 ml = 75.0 ml * (1 L / 1000 ml) = 0.0750 L
moles of HNO3 = 0.0750 L * 0.200 mol/L = 0.0150 mol

Step 3: Use stoichiometry to determine the number of moles of Na+ and NO3- ions:
From the balanced equation, we can see that 1 mol of NaOH reacts with 1 mol of HNO3 to form 1 mol of NaNO3. This means that for each mole of NaOH, we will have 1 mole of Na+ ions and 1 mol of NO3- ions.

Therefore, the moles of Na+ ions remaining in solution = 0.308 mol
And the moles of NO3- ions remaining in solution = 0.0150 mol

c) The resultant solution will be neutral since the balanced equation shows a 1:1 ratio between NaOH and HNO3, yielding NaNO3 and H2O.

For part b, to calculate the concentration of each ion remaining in solution, you'll need to use stoichiometry and the concept of molarity.

First, let's determine the moles of HNO3 present in the solution. We know the volume (75.0 ml) and the molarity (0.200 M), so we can use the formula:

moles of solute = volume of solution (in liters) x molarity

In this case, since the volume is given in milliliters, we need to convert it to liters:

75.0 ml ÷ 1000 = 0.075 L

Now we can calculate the moles of HNO3:

moles of HNO3 = 0.075 L x 0.200 mol/L = 0.015 mol

According to the balanced equation, for every 1 mole of HNO3, there's an equal number of moles of NaOH. Therefore, the moles of NaOH used in the reaction will also be 0.015 mol.

Now, let's find the moles of NaOH remaining in solution. The equation tells us that 1 mole of NaOH reacts with 1 mole of HNO3 to produce 1 mole of NaNO3. Since 0.015 mol of NaOH was used, 0.015 mol of NaOH will react, leaving us with 0 moles of NaOH remaining in solution.

Next, let's consider the product, NaNO3. Since 0.015 mol of NaOH reacts with 0.015 mol of HNO3 to produce 0.015 mol of NaNO3, we have 0.015 mol of NaNO3 in solution.

To calculate the concentration of each ion remaining, we need to consider the dissociation of NaNO3 in water. NaNO3 dissociates into Na+ ions and NO3- ions.

The concentration of Na+ ions in the solution is equal to the number of moles of NaNO3 divided by the volume of the solution in liters:

Concentration of Na+ ions = 0.015 mol ÷ 0.075 L = 0.20 M

Similarly, the concentration of NO3- ions is also 0.20 M.

For part c, since the final concentration of both Na+ and NO3- ions is 0.20 M, which is the concentration of a strong electrolyte, we can conclude that the resultant solution is neutral (neither acidic nor basic).