Calcaulate the pH after 10.0 mL of 0.40 M HCl is added to 20.0 mL of 0.50 M NaOH.

I wasn't sure on how to start this problem. I tried to break the HCl and NaOH apart then I set up ICE for each but I did get the wrong answer of 13.70.

a. 0.40
b. 13.70
c. 13.30
d. 0.30

First you must recognize that HCl is a strong acid and NaOH is a strong base; therefore, the salt produced (NaCl) will not hydrolyze and the solution will be neutral IF the HCl and NaOH EXACTLY neutralize each other. So what you need to do is to calculate the moles HCl and the moles NaOH, see which is in excess and calculate H^+ or OH^- from that. If neither is in excess, the solution will be neutral at pH = 7.

To solve this problem, you can use the concept of neutralization reactions to find the pH.

First, write out the balanced equation for the reaction between HCl and NaOH:
HCl + NaOH -> NaCl + H2O

This reaction produces water and a salt, NaCl. Since NaCl is a strong electrolyte, it completely ionizes in water.

Next, determine the moles of HCl and NaOH present in the solution.

For HCl:
Molarity = 0.40 M
Volume = 10.0 mL = 10.0 cm^3
Convert mL to L: 10.0 cm^3 = 10.0 cm^3 * (1 L / 1000 cm^3) = 0.01 L
Moles = Molarity * Volume = 0.40 M * 0.01 L = 0.004 moles

For NaOH:
Molarity = 0.50 M
Volume = 20.0 mL = 20.0 cm^3
Convert mL to L: 20.0 cm^3 = 20.0 cm^3 * (1 L / 1000 cm^3) = 0.02 L
Moles = Molarity * Volume = 0.50 M * 0.02 L = 0.01 moles

Since NaOH and HCl react in a 1:1 ratio, the moles of HCl consumed will be equal to the moles of NaOH.

Now, calculate the concentration of the remaining NaOH in the solution after the reaction:

Moles of NaOH remaining = Moles of NaOH initially - Moles of NaOH consumed
= 0.01 moles - 0.004 moles = 0.006 moles

Volume of the solution = Volume of HCl added + Volume of NaOH initially = 10.0 mL + 20.0 mL = 30.0 mL = 30.0 cm^3
Convert mL to L: 30.0 cm^3 = 30.0 cm^3 * (1 L / 1000 cm^3) = 0.03 L

Concentration of NaOH remaining = Moles of NaOH remaining / Volume of the solution
= 0.006 moles / 0.03 L = 0.20 M

Finally, calculate the pOH of the solution by taking the negative logarithm of the hydroxide ion concentration:

pOH = -log[OH-]
= -log(0.20)
= 0.70

Since pH + pOH = 14, you can find the pH:

pH + 0.70 = 14
pH = 14 - 0.70
pH = 13.30

Therefore, the correct answer is option c. 13.30.

To solve this problem, you need to understand the concept of neutralization reaction and use the principles of stoichiometry.

1. Write the balanced chemical equation for the reaction between HCl and NaOH:
HCl + NaOH → NaCl + H2O

2. Determine the moles of HCl and NaOH used in the reaction.
Moles of HCl = Molarity of HCl × Volume of HCl solution (in L)
Moles of NaOH = Molarity of NaOH × Volume of NaOH solution (in L)

Moles of HCl = 0.40 M × 0.010 L = 0.004 mol
Moles of NaOH = 0.50 M × 0.020 L = 0.01 mol

3. Determine the limiting reactant. The reactant that is completely consumed in the reaction determines the amount of product formed. In this case, HCl is the limiting reactant because it has fewer moles than NaOH.

4. Use the stoichiometry from the balanced equation to determine the moles of the product formed. Since 1 mole of HCl reacts with 1 mole of NaOH to form 1 mole of water (H2O), the moles of water formed will be equal to the moles of HCl used.

Moles of water formed = 0.004 mol

5. Calculate the concentration of the resulting solution after the neutralization reaction occurs. We need to find the total volume of the resulting solution and then use the moles of water formed to determine the final concentration of H+ or OH-.

Total volume of solution = Volume of HCl solution + Volume of NaOH solution
= 0.010 L + 0.020 L
= 0.030 L

Concentration of H+ or OH- = Moles of water formed / Total volume of solution
= 0.004 mol / 0.030 L
= 0.1333 M

6. Calculate the pH of the solution using the equation:
pH = -log[H+]

pH = -log(0.1333)
pH ≈ 0.13

Therefore, the correct answer is d. 0.30.

It seems that you might have made an error in your calculations or skipped a step, resulting in the wrong answer of 13.70. Double-check your calculations to ensure accuracy.