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chubhunter [2.5K]
3 years ago
6

If 5.0 milliliters of a 0.20 M HCl solution is required to neutralize exactly10. milliliters of NaOH, what is the concentration

of the base?(1) 0.10 M (3) 0.30 M(2) 0.20 M (4) 0.40 M
Chemistry
2 answers:
cricket20 [7]3 years ago
4 0
The answer is (1) 0.1 M NaOH
kirza4 [7]3 years ago
3 0
NaOH+ HCl -> NaCl+ H2O
Based on the above balanced equation, we know that 1 mol HCl= 1 mol NaOH (keep it in mind; it will be used later)

We know that 0.20M HCl solution (soln)= 0.20mol HCl/1L HCl soln (this one is based on the definition of molarity).

First, we have to find the mole of HCl:
5.0mL HCl solution* (1L soln/ 1,000mL soln)* (0.20mol HCl/1L HCl soln)= 1.0*10^(-3)mol HCl.

Now that we know the mole of HCl, let's find the molarity of NaOH:
1.0*10^(-3)mol HCl* (1mol NaOH/1 mol HCl)* (1/10.mL NaOH soln)* (1,000mL NaOH soln/1L NaOH soln)= 0.10mol/L NaOH soln = 0.10M NaOH soln

Hope this would help :))



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It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}

        K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}

Let us assume that the solubility be "s". And, the reaction equation is as follows.

        AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}

     9.84 \times 10^{-21} = s \times s

             s = 9.92 \times 10^{-11}

Also,     Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}

                2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}

                            s = 7.14 \times 10^{-7}

This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the K_{sp} expression as follows.

         K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}

          2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}

       2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}

       [PO^{3-}_{4}]^{2} = 4.88 \times 10^{-24}

                             = 2.21 \times 10^{-12} M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

             AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}

           K_{sp} = [Al^{3+}][PO^{3-}_{4}]

       9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}

                [Al^{3+}] = 4.45 \times 10^{-9} M

Thus, we can conclude that concentration of aluminium will be 4.45 \times 10^{-9} M when calcium begins to precipitate.

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2 years ago
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