An unbuffered solution contaning 0.000018 M HCl .Calculate PH after adding 1 mL of 0.1 M NaOH to 100 mL of solution.

millimoles HCl = mL x M = 100 x 0.000018M = 0.0018.

millimoles NaOH added = 1 mL x 0.1M = 0.1.

..........HCl + NaOH ==> NaCl + H2O
initial..0.0018..0.1.......0......0
change..-0.0018.-0.0018.+0.0018.0.0018
equil....0......0.0982...0.0018.0.0018

You can see that all of the HCl was neutralized and one has a solution of NaCl in excess NaOH.
(OH^-) = 0.0982 mmoles/101 mL = 0.000972M
pOH = -log(OH^-) = -log(0.0.000972 = 3.012
Then pH + pOH = pKw = 1E-14
Solve for pH. It's about 11.
Check my numbers. I've been bad about typos today, especially with the zeros (adding them or omitting them)

23

Kig

Well, well, well, looks like we have a little acid-base titration puzzle here! Let's dive in and have some fun.

First, let's calculate the initial number of moles of HCl in the 100 mL solution.
0.000018 M = moles/volume (L)
Rearranging the formula, we have moles = 0.000018 M x 0.100 L = 0.0000018 moles.

Now, let's calculate the moles of NaOH added.
0.1 M x 0.001 L (that's 1 mL converted to liters) = 0.0001 moles.

Since NaOH and HCl react in a 1:1 ratio, the moles of HCl remaining in solution can be calculated as:
0.0000018 moles - 0.0001 moles = 0.0000017 moles.

Now, we need to calculate the final concentration of HCl in the solution after the reaction. We'll use the final volume of the solution, which is 101 mL (100 mL initial volume + 1 mL of NaOH).

Final concentration = moles/volume = 0.0000017 moles / 0.101 L = 0.0000168 M.

Finally, let's calculate the pH. Since we have an acidic solution, the pH can be calculated using the formula:
pH = -log[H+]

pH = -log(0.0000168) ≈ -(-4.77) (approximating)
pH ≈ 4.77

So, after adding that cheeky 1 mL of NaOH, the pH of this unbuffered solution containing 0.000018 M HCl would be approximately 4.77. Ta-da!

To calculate the pH after adding 1 mL of 0.1 M NaOH to a 100 mL solution containing 0.000018 M HCl, you need to use the concept of stoichiometry, which involves balancing the chemical equation and calculating the resulting concentrations of the products.

First, let's write the balanced chemical equation for the neutralization reaction between HCl and NaOH:

HCl + NaOH -> NaCl + H2O

We can see that one mole of HCl reacts with one mole of NaOH to form one mole of NaCl and one mole of water.

Next, let's calculate the moles of HCl in 100 mL of the solution containing 0.000018 M HCl:

moles of HCl = concentration x volume
= 0.000018 M x 0.100 L
= 1.8 x 10^-6 moles

Since the stoichiometry of the reaction is 1:1, the moles of NaOH needed to neutralize the HCl is also 1.8 x 10^-6 moles.

Now, let's calculate the resulting concentration of HCl after the NaOH is added and neutralizes the HCl:

volume of solution = 100 mL + 1 mL = 101 mL = 0.101 L

concentration of HCl after neutralization = moles of HCl / volume of solution
= (1.8 x 10^-6 moles) / 0.101 L
= 1.782 x 10^-5 M

Finally, we can calculate the pOH and pH of the solution using the concentration of HCl after neutralization:

pOH = -log[OH-] = -log(0.1 M / 1000) = -log(0.0001)

pH = 14 - pOH

To calculate the final pH, you need to know the pOH value or the concentration of hydroxide ions (OH-) in the NaOH solution used for neutralization. The provided information does not indicate this concentration, so you cannot determine the exact pH value without that additional information.

Please provide the concentration of hydroxide ions in the NaOH solution or any other relevant information so that we can calculate the final pH.