Question

Given log (0.00001)=-5; log(0.0001)=-4; and log(0.001)=-3. If you added a drop (about 0.05 mL) of 1.0 M HCL to one liter of pure water (assume pH 7.0), the pH would become:

a) 6.5
b) 5.5
c) 4.5
d) 3.5

Answer 1

Adding 0.05 mL of 1.0 M HCl to one liter of water will result in a new H+ concentration of about 5.0 × 10−5 M and a pH close to 4.0, making the correct choice (d) 3.5.

Step-by-step explanation:

To determine the change in pH after adding 0.05 mL of 1.0 M HCl to one liter of pure water with an initial pH of 7.0, we need to calculate the concentration of hydrogen ions after the addition and then find the new pH.

First, let's find out how many moles of HCl are in 0.05 mL:

0.05 mL of HCl is equivalent to 0.05/1000 = 0.00005 L.

Since the concentration of HCl is 1.0 M, the number of moles in 0.05 mL is 1.0 mol/L × 0.00005 L = 5.0 × 10−5 moles.

Adding these moles to 1 liter of water will dilute the concentration:

The total volume of the solution is now approximately 1.00005 L.

The new concentration of H+ is 5.0 × 10−5 moles / 1.00005 L ≈ 5.0 × 10−5 M.

Now, we'll calculate the pH:

pH = -log[H+] = -log(5.0 × 10−5)

Looking at the given logs, we can approximate this to be close to 4.

Therefore, the pH of the water after adding a drop of HCl would be approximately 4.0, not taking into account the significant figure considerations. This would make the correct choice (d) 3.5.