Sodium oxide, reacts with water to give NaOH. A) Write a balance equation? B)What is the pH of the solution prepared by allowing 1.55 g of Na2O to react with 500.0 ml of water? (assume there is no volume change) C) How many milliliters of 0.0100M HCL are needed to neutralize the NaOH prepared in(b)

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A.

Na2O + HOH ==> 2NaOH

B. There is a long way and a short way of doing this. The long way is to make a stoichiometry problem out of it, convert g Na2O to moles, calculate moles NaOH formed, then use M = moles/L.
A shorter way is to forget about stoichiometry and reason it out.
moles Na2O = 1.55 g/molar mass Na2O = ??
moles NaOH = 2*moles Na2O
M = moles/L
M NaOH = molarity of OH^-
Then pOH = -log(OH^-) and
pH + pOH = pKw = 14
Solve for pH.

C. Use
mL acid x M acid = mL base x M base.
Solve for the unknown, mL acid.

A) The balanced equation for the reaction between sodium oxide and water is:

Na2O + H2O -> 2 NaOH

B) To find the pH of the resulting solution, we need to first calculate the concentration of NaOH.

First, we need to convert grams of Na2O to moles:
1.55 g Na2O * (1 mol Na2O / 61.98 g Na2O) = 0.0250 mol Na2O

Next, we need to calculate the concentration of NaOH in the solution:
Volume of solution = 500.0 mL = 0.500 L

Concentration of NaOH = moles of NaOH / volume of solution
Concentration of NaOH = (0.0250 mol Na2O * 2 mol NaOH / 1 mol Na2O) / 0.500 L
Concentration of NaOH = 0.100 M

The pH of a 0.100 M NaOH solution can be calculated using the equation:

pOH = -log10[OH-]
pOH = -log10(0.100) = 1

Since pOH + pH = 14, we can find the pH:
pH = 14 - pOH = 14 - 1 = 13

Therefore, the pH of the solution is 13.

C) To calculate the volume of 0.0100 M HCl needed to neutralize the NaOH solution, we can use the equation:

M1V1 = M2V2

Where:
M1 = concentration of HCl
V1 = volume of HCl needed
M2 = concentration of NaOH
V2 = volume of NaOH solution

Rearranging the equation, we get:
V1 = (M2 * V2) / M1

Plugging in the values:
M1 = 0.0100 M
M2 = 0.100 M
V2 = 0.500 L

V1 = (0.100 M * 0.500 L) / 0.0100 M
V1 = 5.00 L

Therefore, 5.00 L (or 5000 mL) of 0.0100 M HCl solution is needed to neutralize the NaOH solution.

A) To write a balanced equation for the reaction of sodium oxide (Na2O) with water (H2O) to give sodium hydroxide (NaOH), we need to ensure that the number and type of atoms are the same on both sides of the equation.

Na2O + H2O → 2NaOH

B) To find the pH of the solution prepared by allowing 1.55 g of Na2O to react with 500.0 ml of water, we need to determine the concentration of NaOH in the solution.

First, let's convert the given mass of Na2O to moles:
1 mol Na2O = 61.98 g Na2O
1.55 g Na2O × (1 mol Na2O / 61.98 g Na2O) = 0.025 mol Na2O

Next, we need to calculate the concentration of NaOH in the solution. Since the balanced equation tells us that 1 mole of Na2O yields 2 moles of NaOH, the number of moles of NaOH is twice the number of moles of Na2O:

0.025 mol Na2O × (2 mol NaOH / 1 mol Na2O) = 0.050 mol NaOH

Now, let's calculate the concentration of NaOH in moles per liter (Molarity):
Volume of solution = 500.0 ml = 0.500 L
Concentration of NaOH = 0.050 mol NaOH / 0.500 L = 0.100 M NaOH

Since NaOH is a strong base, fully dissociating in water, we can assume that the concentration of hydroxide ion (OH-) is also 0.100 M.

To find the pH of the solution, we can use the equation: pH = 14 - pOH

pOH = -log[OH-]
pOH = -log(0.100)
pOH ≈ 1

pH = 14 - 1
pH ≈ 13

Therefore, the pH of the solution is approximately 13.

C) To find the volume of 0.0100 M HCl needed to neutralize the NaOH prepared in part B, we can use stoichiometry.

From the balanced equation in part A, we know that 1 mol of NaOH reacts with 1 mol of HCl.

The number of moles of NaOH is:
0.050 mol NaOH

Therefore, we need the same number of moles of HCl:
0.050 mol HCl

To calculate the volume of 0.0100 M HCl needed, we can use the Molarity equation:
Molarity (M) = Moles / Volume (L)

0.0100 M = 0.050 mol / Volume
Volume = 0.050 mol / 0.0100 M ≈ 5.0 L

Therefore, approximately 5.0 mL of 0.0100 M HCl is needed to neutralize the NaOH prepared in part B.