What mass of NaOH pellets is required to prepare 3.5 L of 0.50 M NaOH solution?

Question

What mass of NaOH pellets is required to prepare 3.5 L of 0.50 M NaOH solution? How would you prepare 275 mL of 0.350 M NaCI solution using an available 2.00 M solution? Many metal ions are precipitated from solution by the sulfide ion. As an example, consider treating a solution of copper (II) suffato with sodium sulfide solution: CuSO_4(aq) + Na_2S_(aq) rightarrow CuS_(s) + Na_2SO_4(aq) What volume of 0.105 M Na_2 S solution would be required to precipitate all of the copper (II) ion from 27 5 ml of 0.121 M CuSO_4 solution? What volume of 0 502 M NaOH solution would be required to neutralize 27.2 mL of 0 491 MHNO_3 solution.

Explanation / Answer

5)

Number of Moles = Molarity (M) * Volume of NaOH(L)

=> 0.50M * 3.50L

=> 1.75 moles

Molar mass of NaOH = 23 + 16 + 1 = 40 gm/mol

Mass of NaOH pallets required = 40 gm/mol * 1.75 mol = 70 grams

6)

M1V1 = M2V2

(275mL)*0.350 = (2M) * VmL

V = 48.125 mL

Add (275-48.125)mL of water in 48.5mL of 2M NaCl solution

7)

Number of moles of CusO4 = Number of moles of Na2S

27.5 * 0.121 = V(mL) * 0.105

V(mL) = 31.690 mL

8)

Number of moles of HNO3 = Number of moles of NaOH

27.2 * 0.491 = 0.502 * V(mL)

V(mL) = 26.603 mL

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