pH of a diluted with distilled water

[christotaku]

Hi!
This is my first post here so I am sorry if I am not doing something right.

My question is how to measure pH of a solution that results when a substance is diluted with distilled water. Here in Bulgaria I don't remember solving such problems in the chemistry class in school but I need it for an exam, so I decided to ask here.

Examples:
What is the pH of the solution that results when 10ml of 10^-5mol/l HCl is diluted to 10l with distilled water?
What is the pH of the solution that results when 10ml of 10^-3mol/l NaOH is diluted to 1l with distilled water?
What is the pH of the solution that results when 10ml of 10^-2mol/l CH₃COOH is diluted to 1l with distilled water?
What is the pH of the solution that results when 10ml of 10^-3mol/l H₂SO₄ is diluted to 1l with distilled water?

[engineer7]

1) "... how to measure pH..." You could use pH testers.

2) Depending on the level of knowledge you are being tested for, "distilled water" may mean pH 7 or lower, say pH 5 - pH 6,5.

3) What is the pH of the solution that results when 10ml of 10^-5mol/l HCl is diluted to 10l with distilled water

For water with pH₇.

3.1 Finding initial number of moles of H+
a) converting ml into L 10mL=0.01L.
b) using molar concentration 10^-5 mol/L, calculating 0.01L x 10^-5 mol/L= 10^-7 moles of H+ was in the initial stock

3.2. to find number of H+ moles after diluting with water
c) volume of water added 10L - 0.01L= 9.99L
d) 9.99L of water with pH₇ contains 9.99x 10^-7 moles of H+
e) resulting solution contains (initial+added) moles of H+: 1 x 10^-7 moles + 1x10^-7 x 9.99 moles=10.99x10^-7M or 1.099x10^-6 moles

3.3. Calculating pH.
f) finding molar concentration, considering total volume 10L: 1.099*10^-6 / 10L= 1.099x10^-7
g) pH is negaive logarithm: -lg 1.099x10^-6 = 6.96 (or, appr 7)

[christotaku]

1) "... how to measure pH..." You could use pH testers.

2) Depending on the level of knowledge you are being tested for, "distilled water" may mean pH 7 or lower, say pH 5 - pH 6,5.

3) What is the pH of the solution that results when 10ml of 10^-5mol/l HCl is diluted to 10l with distilled water

For water with pH₇.

3.1 Finding initial number of moles of H+
a) converting ml into L 10mL=0.01L.
b) using molar concentration 10^-5 mol/L, calculating 0.01L x 10^-5 mol/L= 10^-7 moles of H+ was in the initial stock

3.2. to find number of H+ moles after diluting with water
c) volume of water added 10L - 0.01L= 9.99L
d) 9.99L of water with pH₇ contains 9.99x 10^-7 moles of H+
e) resulting solution contains (initial+added) moles of H+: 1 x 10^-7 moles + 1x10^-7 x 9.99 moles=10.99x10^-7M or 1.099x10^-6 moles

3.3. Calculating pH.
f) finding molar concentration, considering total volume 10L: 1.099*10^-6 / 10L= 1.099x10^-7
g) pH is negaive logarithm: -lg 1.099x10^-6 = 6.96 (or, appr 7)

I think I got that but what do we do with

Code: Select all

CH3COOH

for example, when I calculated it I got appr 4 but the answer is different. And do we multiply when there are more H+, for example in

Code: Select all

H2SO4

[engineer7]

"...what do we do..." I guess, first we to thank people for helping us: "благодарÑ!" ))

Second, CH₃COOH is a weak acid, therefore you, pls look for "acid ionization constant Ka" (should be provided).

Third, H₂SO₄ - even "worse" )), as it dissociates in 2 steps:
H + HSO₄ (complete),
then H dissociates to H + SO₄ (not complete, so Ka involved), so, strictly speaking you can't multiply by 2 unless your teacher allows this.

[christotaku]

"...what do we do..." I guess, first we to thank people for helping us: "благодарÑ!" ))

Second, CH₃COOH is a weak acid, therefore you, pls look for "acid ionization constant Ka" (should be provided).

Third, H₂SO₄ - even "worse" )), as it dissociates in 2 steps:
H + HSO₄ (complete),
then H dissociates to H + SO₄ (not complete, so Ka involved), so, strictly speaking you can't multiply by 2 unless your teacher allows this.

I am terribly sorry. Thanks a lot! "Ð‘Ð»Ð°Ð³Ð¾Ð´Ð°Ñ€Ñ Ð¼Ð½Ð¾Ð³Ð¾!" if you prefer Bulgarian. So with acids we have to keep in mind Ka.
Thank you, again. It was very rude of me.