standard solutions of calcium ion used to test for water hardness are prepa calc

Question

standard solutions of calcium ion used to test for water hardness are prepa calcium carbonate, CaCO3, in dilute hydrochloric acid. A 1.762 9 sam mL volumetric flask and dissolved in HCI. Then the solution is di volumetric flask. Calculate the resulting molarity of calcium ion are prepared by dissolving pure 1.762 g sample of CaCOs is placed in a 200. mark of the diluted to the calibration 7. -/1 pointsZumintro6 15.P.060. Ir 55 mL of a 0.187 M NaOH is diluted to a final volume of 125 ml, what is the diluted solution? 8. -/1 pointsZumIntro6 15.P.O65. Many metal ions are precipitated from solution by the sulfide ion. solution of copper(II) sulfate with sodium sulfide solution. CuSO4(aq) + Na2S(aq) CuS(s) + Na2SO4(aq) As an example, consider treating a (II) ion from What volume of 0.118 M Na2s solution would be required to precipitate all of the copper 27.6 mL of 0.138 M CuSO4 solution? -/4 pointsZumintro6 15.P.073. hat volume of 1.25 M NaOH is required to neutralize each of the following solutions? (a) 25.9 mL of 0.172 M acetic acid, HC2H302 (b) 37.1 mL of 0.123 M hydrofluoric acid, HF (c) 13.3 mlL of 0.139 M phosphoric acid, H3PO mL (d) 36.5 mL of 0.220 M sulfuric acid, H2S04 ebassign.net/web/Student/Assignment-Responses/last?dep-16951685

Explanation / Answer

Answer:

6) Given weight of CaCO3=1.762 g, and molar massof CaCO3=100 g/mol.

This solution was made in 200 mL volumetric flask by using HCl, so volume of solution=200 mL=0.2 L (1 L=1000 mL).

We know the concentration is equall to moles of solute per 1L of solvent.

Here moles of CaCO3=mass/molar mass=1.762 g/100g/mol=0.01762 mol.

So concentration=moles/volume=0.01762 mol/0.2 L=0.0881 M.

The concentration of CaCO3=0.0881 M.

In CaCO3, the mole ratio between CaCO3 and Ca+2 ion is 1:1,

So the molarity of Ca+2 ion=0.062 mol/0.2 L=0.0881 M.

7) Given molarity of NaOH=0.187 M

Now this solution was diluted to 125 mL by adding 55 mL of 0.187 M NaOH.

By dilution law N1V1=N2V2

Here N1=0.187 M=0.187 N (since for NaOH gram molecular weight=gram equivalent weight), V1=55 mL,

V2=125 mL and N2=?

Therefore N2=N1V1/V2=(0.187 Mx55 mL)/125 mL=0.08228 M.

So the diluted NaOH concentration is 0.08228 M.

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