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Titration qu (1 Viewer)

VenomP

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In a titration, 25.0 mL of sodium hydroxide solution was pipetted into a conical flask. A 0.100 mol/L solution of sulfuric acid was run from the burette into the flask. An indicator in the flask changed colour when 22.0 mL of the acid had been added.

a) Write a balanced equation for the H reaction.

b) Calculate the number of moles of sulfuric acid added.

c) Calculate the number of moles of NaOH reacted.

d) What is the molarity of the NaOH solution?

e) What indicator could be used in this titration? Why?

I think I've done okay up til part d, where I have no idea what to do next. However I could be completely off the money.

Sorry, but I don't have answers for these questions.

Could someone please explain the processes to me?
 
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Use:
cava/a = cbvb/b
where ci is the concentration of i, vi is the volume of i, and i is the stoichiometric coefficient.

VenomP said:
What indicator could be used in this titration? Why?
Anyyything, as the pH change is large enough to cover the colour changes for all standard indicators.

Just remember, when deciding what to use, consider the products of the titration, and consider how those substances, when dissociated in water, will affect the pH of the water.
Weak a. and weak base: nothing; cannot titrate.
W. a, st. b.: Phen.
St. a., w. b.: Methyl red/orange.

Despite how I was taught in school, add acids to bases because acids do not tend to form deposits on the delicate equipment.
Then again, all the equipment is coated with this thingo that I forget now (that you don't need to know for the sake of the syllabus).
 

VenomP

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Ah okay brilliant. Thanks. We hadn't learned that cava/a = cbvb/b formula yet, so I'm not quite sure what the stoichiometric coefficient is at the moment. But thanks anyway.

Does this formula apply to all the questions? Could you possibly type out the working for me? I'm having a little trouble applying it...
 
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Sorry. I think I should just clearly answer and explain all.
VenomP said:
In a titration, 25.0 mL (volume of base) of sodium hydroxide solution (base) was pipetted into a conical flask. A 0.100 mol/L (conc. of acid) solution of sulfuric acid (acid) was run from the burette into the flask. An indicator in the flask changed colour when 22.0 mL (volume of acid) of the acid had been added.
We hadn't learned that cava/a = cbvb/b formula yet
Surely you've been taught some variation of it?

As far as I know, there are at least three different formulae - or, three ways of writing - for titrations; the underlying concept is that x moles of acid react with y moles of base, n=cv, and (the law of conservation of mass).

The formula I've presented you with works for all titrations, and IMO it's not that confusing.
I'm not quite sure what the stoichiometric coefficient is at the moment
"When you balance your equation, how many moles of this and that do you need to use?"

The integral coefficient of each substance is the stoichiometric coefficient of that respective substance.

For example, here:
NaOH + H2SO4 --> H2O + Na2SO4

Balance:
2NaOH + (1)H2SO4 --> 2H2O + NA2SO4

The (stoichiometric) coefficients here are, for base and acid respectively, 2 and 1; these two values are your 'b' and 'a' in that formula I presented you.
Could you possibly type out the working for me? I'm having a little trouble applying it...
cava/a = cbvb/b

.1 x 22 / 1 = 25 x x / 2

Then, make your unknown concentration/molarity of your base, x, your subject.

According to mental arithmetic, .176 is your cb. Correct me if I'm wrong.
 
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brenton1987

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In a titration, 25.0 mL of sodium hydroxide solution was pipetted into a conical flask. A 0.100 mol/L solution of sulfuric acid was run from the burette into the flask. An indicator in the flask changed colour when 22.0 mL of the acid had been added.
a) Write a balanced equation for the H reaction.
b) Calculate the number of moles of sulfuric acid added.
c) Calculate the number of moles of NaOH reacted.
d) What is the molarity of the NaOH solution?
e) What indicator could be used in this titration? Why?
2 NaOH + H2SO4 --> Na2SO4 + 2 H2O

CH2SO4 = 0.100 mol L-1
VH2SO4 = 0.0220 L

nH2SO4 = 0.0022 mol

nNaOH = 0.0044 mol
VNaOH = 0.025 L
CNaOH = 0.176 mol L-1

Strong acid/Strong base titration results in a neutral solution. Bromothymol blue (6.0–7.6)
 

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