Question

Calculate the number of milliliters of 0.587 M NaOH required to precipitate all of the Ni2+ ions in 163 mL of 0.445 M NiBr2 solution as Ni(OH)2. The equation for the reaction is: NiBr2(aq) + 2NaOH(aq) Ni(OH)2(s) + 2NaBr(aq) mL NaOH Submit AnswerRetry Entire Group

Answer 1

Answer:

We need 247 mL of NaOH

Explanation:

Step 1: Data given

Molarity of NaOH = 0.587 M

Volume of 0.445 M NiBr2 solution = 163 mL = 0.163 L

Step 2: The balanced equation

NiBr2(aq) + 2NaOH(aq) Ni(OH)2(s) + 2NaBr(aq)

Step 3: Calculate moles of NiBr2

Moles NiBR2 = Molarity NiBR2 * volume

Moles NiBR2 = 0.445 M * 0.163 L

Moles NiBR2 = 0.0725 moles

Step 3: Calculate moles of NaOH

For 1 mol NiBr2 consumed, we need 2 moles NaOH

For 0.0725 moles NiBR2, we need 2* 0.0725 = 0.145 moles NaOH

Step 4: Calculate volume of NaOH

Volume = moles NaOH / Molarity NaOH

Volume = 0.145 moles / 0.587 M

volume = 0.247L = 247 mL

We need 247 mL of NaOH