a 10ml sample of a HCl solution has a pH of 2, what volume of water must be added to change the pH to 4.0?

Answer 1

Answer: 990 mL water


Explanation:

Remember that pH is simply the negative log of the Hydrogen Ion concentration.

pH = -log [H+]

So what you've started out with is:

2 = - log [H+]

Solve for original [H+]:

[H+] = 10^-2
[H+] = 0.010 Molarity


Determine the [H+] of a pH 4.0 solution

4 = -log[H+]

[H+] = 0.00010 Molarity


Now go back to the first calculation you made:

You have an original molarity of 0.01 moles/l and 0.01 Liters (10mL).

Multiply across to determine the moles of H+ you have

0.01 (moles/L) X 0.01 Liters = 0.00010 moles H+


Now use these moles to figure out the new volume needed for a pH 4.0 solution.


[H+] molarity needed is 0.00010 moles/L

You already have that many moles of H+, so you simply need 1L of water. Since you already have 10 mL in the original solution, you need 990 mL of water.

Answer 2

Answer:

pH = - log [ H+]

- log [ H+] = 2 ; [H+] = antilog (- 2 ) = 10*-2 mol L*-1

So concentration of H+ = 10*-2 mol L*-1

Next by the same procedure we can determine the concentration
of H+ that gives a pH of 4 which is 10*-4 mol L*-1.

Using the latter concentration we can find out in what volume of solution would contain 10*-2 mol H+. Why? Because when you add water to the 10 ml of HCl solution you would not be changing the amount of mol of H+ ions ( 10*-2 mol ) in it.

10*-4 mol is contained in 1000 ml of solution
therefore
10*-2 mol is contained in (10*-2 x 1000)/ 10*-4

= 100000 ml or 1x 10*5 ml

The difference between this calculated volume and 10 ml is the volume of water needed to be added.

Water that must be added is 100000 - 10 = 99990 ml

Answer 3

the moles of cation H+ in the first solution = 0.01
the moles of cation H+ in the second solution = 10^(-4)
to reach the pH of 4 the solution must have less moles of cation H+, the amount of cation H+ must be subtracted = 0.01 - 10^(-4) = 9.9*10^(-3)
it is also the amount of anion OH- that we have to add to the first solution by adding water , so we have the moles of water = moles of OH- = 9.9*10(-3) the volume of water that will be added is 1*18*9.9*10(-3) = 0.1782 ml

Answer 4

Find the molar concentration first. Take the inverse log of -2 (as pH is -log(molarconcentration)), and then set up an algebraic equation of molar concentrations

Molar concentration x .01 mL = moles of H+

x = volume of water total.

Moles of H+/(.01) + 0/(x-.01) = moles of H+/(x) such that -log(moles of H+/(x+y)) is equal to four.