If 20 mL of 0.01 M aqueous HCl is required to neutralize 30 mL of an aqueous solution of NaOH, determine the molarity of the NaOH solution.

==> I got 0.01 M NaOH as my answer. Is this correct?

No, it isn't correct although I suspect you worked it correctly. I suspect you simply read your calculator incorrectly.

mL x M = mL x M (you can use this simplified expression since the reaction is 1:1 and ONLY then).
20 x 0.01 = 30 x M
M = (20*0.01)/30 = 0.00667 which rounds to 0.0067 M.
If your calculator is like most, you have it set to read to two places, so the calculator rounds 0.006 to 0.01.

Ohhh I see. This makes more sense. Thank you! :)

Well, aren't you just the mighty neutralizer! Unfortunately, I must inform you that your answer is not correct. Remember, when an acid and a base neutralize each other, the number of moles of hydrogen ions (H+) from the acid must be equal to the number of moles of hydroxide ions (OH-) from the base.

Since the HCl solution is 0.01 M and you used 20 mL of it, we can calculate the number of moles of HCl using the formula: Moles = Molarity x Volume

So, Moles of HCl = 0.01 M x 0.020 L = 0.0002 moles

Since HCl and NaOH react in a 1:1 ratio, the number of moles of NaOH in the solution is also 0.0002 moles.

Now, we can use the formula Molarity = Moles / Volume to find the molarity of NaOH.

Molarity of NaOH = 0.0002 moles / 0.030 L ≈ 0.0067 M

So, the molarity of the NaOH solution is approximately 0.0067 M, not 0.01 M. Keep those calculations flowing!

Let me know if you need any more help or if you just want to hear another pun!

To determine the molarity of the NaOH solution, we can use the concept of stoichiometry.

Step 1: Identify the balanced chemical equation for the reaction between HCl and NaOH. In this case, it is:

HCl + NaOH → NaCl + H2O

Step 2: Determine the number of moles of HCl used. We can use the formula:

moles = volume (in liters) x molarity

Given that the volume of HCl used is 20 mL (0.02 L) and the molarity is 0.01 M, we can calculate the moles of HCl used:

moles of HCl = 0.02 L x 0.01 M = 0.0002 moles

Step 3: Based on the balanced chemical equation, we can see that one mole of HCl reacts with one mole of NaOH to produce one mole of NaCl. Therefore, the moles of NaOH used will be equal to the moles of HCl used.

moles of NaOH = 0.0002 moles

Step 4: Calculate the molarity of the NaOH solution using the formula:

molarity = moles / volume (in liters)

Given that the volume of the NaOH solution is 30 mL (0.03 L), we can calculate the molarity of the NaOH solution:

molarity of NaOH = 0.0002 moles / 0.03 L = 0.00667 M

Hence, the molarity of the NaOH solution is 0.00667 M, which is not equal to 0.01 M.

To determine the molarity of the NaOH solution, we can use the concept of stoichiometry and the equation for neutralization:

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

2. Determine the mole ratio between HCl and NaOH from the balanced equation.
From the balanced equation, we can see that the mole ratio of HCl to NaOH is 1:1.

3. Calculate the number of moles of HCl.
We are given the volume of HCl solution as 20 mL (0.020 L) and the molarity of HCl as 0.01 M.
Moles of HCl = Volume (in L) × Molarity
= 0.020 L × 0.01 M
= 0.0002 moles

4. Since the mole ratio of HCl to NaOH is 1:1, the number of moles of NaOH will also be 0.0002 moles.

5. Calculate the molarity of NaOH solution.
We are given the volume of NaOH solution as 30 mL (0.030 L).
Molarity of NaOH = Moles of NaOH / Volume (in L)
= 0.0002 moles / 0.030 L
= 0.0067 M

Therefore, the molarity of the NaOH solution is 0.0067 M, not 0.01 M.