Question

1) What mass of pure CaCO3 must be used to prepare 500 mL of an aqueous solution that has a concentration of 950 mg/L Ca2+ ion?

2)A 10 mL pipet full of solution containing .0450 M NaCl, .0545 M HCl, .0255 M MgCl2, and .0380 M ZnCl2 is put into a column of strong cation exchange resin which is in its hydrogen cycle or form. The eluted H+ required 19.10 mL of NaOH for complete neutralization. What is the molarity of the NaOH solution?

Answer 1

1.) 950 mg /L means

For 1 L solution we need 950mg Ca+2 ions

For 500mL solution we need = 950/2 = 475 mg of Ca+2 ions

One mole of Ca+2 = 40 g

(1g = 1000 mg)

For 0.450 g of Ca+2 = 0.450 / 40 = 0.01125 moles

1 mole of CaCO3 gives = 1 mole of Ca+2

So for

0.01125 moles of Ca+2 we needs = 0.01125 moles of CaCO3

MW of CaCO3 = 100.08 mg

0.01125 moles of CaCO3 weighs = 0.01125 × 100.08

= 1.1259

Which is approximately 1.13 gm of CaCO3

Please ask each question separately unless they are not related to each other.

answered by: anonymous