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tiny-mole [99]
2 years ago
9

If 15.0 mL of phosphoric acid completely neutralizes 38.5 mL of 0.150 mol/L calcium hydroxide, what is the concentration of the

phosphoric acid?
Chemistry
1 answer:
Sedbober [7]2 years ago
7 0

Answer:

Let me give it a try.

H3PO4 + Ca(OH)2 = Ca3(PO4)2 + H2O

Balancing this reaction

2H3PO4 + 3Ca(OH)2 == Ca3(PO4)2 + 6H2O.

Moles= Molarity x Volume

Volume = 38.5ml = 0.0385L

Moles of Ca hydroxide = 0.150m/L x 0.0385L

(Notice the units canceling out...leaving moles).

=0.005775moles of Ca(OH)2.

From balanced reaction...

3moles of Ca(OH)2 completely reacts with 2moles of H3PO4

0.005775moles of Ca(OH)2 would completely react with....

= 0.005775 x 2/(3)

=0.00385moles of H3PO4.

Now we're looking for its Concentration in Mol/L

Molarity=Moles of solute/Volume of solution(in L)

Volume of solution assuming no other additions to the reaction = 15ml + 38.5ml =53.5ml =0.0535L

Molarity = 0.00385/0.0535

=0.072Mol/L.

If this is wrong

then Simply Try The formula for Mixing of solutions

C1V1 = C2V2

0.15 x 38.5 = C2 x (15+38.5)

C2 = 0.11M/L.

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Answer:

0.1056 mole

Explanation:

As Sally knows that the charge on the metal ion is n = +2

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The volume of the $MCl_n$ taken in that compartment = 6.380 mL

So, the number of moles of $M^{2+} = 8.279 \times 6.380$

                                                      = 52.82 m mol

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$MCl_n \rightarrow M^{n+}+nCl^-$

But n = 2

Therefore, moles of $Cl^-$ = 2 x moles of $M^{n+}$

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The solubility of magnesium phosphate at a given temperature is 0.173 g/L. Calculate the Ksp at this temperature. After you calc
jek_recluse [69]

Answer: K_{sp}=1.25\times 10^{-14}

pK_{sp}=13.90

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as K_{sp}  

The equation for the ionization of magnesium phosphate is given as:

Mg_3(PO_4)_2\rightarrow 3Mg^{2+}+2PO_4^{3-}

 When the solubility of Mg_3(PO_4)_2 is S moles/liter, then the solubility of Mg^{2+} will be 3S moles\liter and solubility of PO_4^{3-} will be 2S moles/liter.

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