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Chemistry 15 Online
OpenStudy (ashley1nonly):

An impure sample containing phosphoric acid was completely neutralized with strong base. The titration required 90.0M NaOH solution. calculate the equivanent weight of the acid how many moles of protons were titrated how many moles and grams of H3PO4 were titrated

OpenStudy (anonymous):

I think we need more info, like V of NaOH

OpenStudy (ashley1nonly):

Sorry 90.0ml of 0.222 M NaOH

OpenStudy (anonymous):

Have you solved the first problem?

OpenStudy (ashley1nonly):

no

OpenStudy (anonymous):

Ok, first calculate the moles of NaOH. Then apply it into the equation \(3NaOH+H_3PO_4\rightarrow Na_3PO_4+3H_2O\) 3 moles 1 moles n moles ? moles \(n_{H_3PO_4}=n_{NaOH}\times 1 \div 3\)

OpenStudy (ashley1nonly):

how do we get the N values

OpenStudy (ashley1nonly):

0.222=x ----- .09L for moles

OpenStudy (ashley1nonly):

moles= 0.0198moles NaOh

OpenStudy (ashley1nonly):

could we use the molarity M= moles/L 0.22=moles/.09

OpenStudy (anonymous):

Sorry, inappropriate formula

OpenStudy (anonymous):

Yes. That one

OpenStudy (anonymous):

And by the way, we are solving #3, sorry

OpenStudy (ashley1nonly):

ok

OpenStudy (anonymous):

For the first one, \(\Large equivalent~weight =\frac{molar~mass}{numbers~of~H}\)

OpenStudy (ashley1nonly):

we are using the molar mass of H3PO4 for that one or NaOH

OpenStudy (ashley1nonly):

97.963g H3PO4 ------------- 3 for (h3)

OpenStudy (anonymous):

\(H_3PO_4\). We can use that formula to any acids. For bases, we use \(\Large equivalent~ weight=\frac{molar ~mass}{numbers ~of ~OH}\)

OpenStudy (ashley1nonly):

=32.654

OpenStudy (anonymous):

Correct

OpenStudy (ashley1nonly):

okay for the 3rd one we have 0.0198moles NaOh where do we go from that to get the moles or was that it

OpenStudy (ashley1nonly):

we have to relate that using the balanced equation

OpenStudy (anonymous):

Yes. Based on the equation, 3 moles of NaOH will react with 1 moles of H3PO4. Then 0.0198 moles of NaOH will react with....

OpenStudy (ashley1nonly):

0.0198 moles NaOh 1 mole H3PO4 ------------------ X----------------= .0066 moles H3PO4 1 3 moles NaOH .0066 moles H3PO4 97.963g H3PO4 ------------------ x---------------- = .64656 g H3PO4 1 1 mole H3PO4

OpenStudy (anonymous):

Exactly. Good job

OpenStudy (ashley1nonly):

how do we find the moles of protons

OpenStudy (anonymous):

Actually, protons in this problem is \(H^+\). As you know, \(H\) has 1 p and 1 e. When it is \(H^+\), there is 1 p left

OpenStudy (anonymous):

All you need to do is to find the moles of \(H^+\)

OpenStudy (ashley1nonly):

in NaOH or H3PO4

OpenStudy (anonymous):

Only acids produce \(H^+\). Bases produce \(OH^-\)

OpenStudy (ashley1nonly):

so we will use the weak acid H3PO4 .0066 moles H3PO4 3 moles H ----------------- x -----------------=.0198 1 1 mole H3PO4

OpenStudy (ashley1nonly):

.0198 moles H

OpenStudy (anonymous):

Right

OpenStudy (anonymous):

Well done

OpenStudy (ashley1nonly):

and that equals the number of my protons

OpenStudy (ashley1nonly):

or should i take it to grams

OpenStudy (anonymous):

The moles of your protons :) No, the problem asks for the moles of protons, not its mass

OpenStudy (ashley1nonly):

oh okay thank you very much for all of your help, its sometimes hard to relate all the knowledge into one problem

OpenStudy (anonymous):

You are welcome. And yes, it is

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