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?
pH of a diluted with distilled water
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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)
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)
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- Jr. Member
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I think I got that but what do we do withengineer7 wrote: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)
Code: Select all
CH3COOH
Code: Select all
H2SO4
pH, diluted acids
"...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.
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.
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- Jr. Member
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- Joined: Fri May 13, 2016 2:57 am
Re: pH, diluted acids
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.engineer7 wrote:"...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.
Thank you, again. It was very rude of me.