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Calculating The Mass of Zn(OH)2 That Reacted With HBr

October 9th, 2008

Chemistry Help - Chemistry Problem

A solid White sample of Zn(OH)2 is added to 0.300 L of 0.420 M  HBr. solution. The solution that remains is still acidic and It is then titrated with 0.420 M NaOH solution, and it takes 94.5 mL of the NaOH solution to reach the equivalence point.

What mass of Zn(OH)2 was added to the HBr solution?

Chemistry Tutor Answer

To solve this problem you only need to find how many moles of HBr that reacted with Zinc hydroxide Zn(OH)2 How you will find it? First you must find the initial moles of HBr and than subtracts with the moles of HBr that reacted with natrium hydroxide.

After that change the mol of HBr into moles of Zn(OH)2 using the coefficient of the reaction and multiply with the molecular mass to obtain the mass of zinc hydroxide. here are the problem solving of that chemistry problem.

moles of HBr initial
= 0.300 L x 0.420
= 0.126 mol

mol of NaOH
= 0.420 M x 94.5 mL
= 39.69 mmol
= 0.039 mol

NaOH reacted with HBr according this reaction, remember to find the mol of HBr you only need to compare the reaction coefficient since the coefficient of HBr and NaOH are equal thus their moles would be the same.

NaOH + HBr -> NaBr + H2O

from the reaction the mol of HBr are
= 1/1 x 0.039 mol
= 0.039 mol

thus the mol of HBr that reacted with Zn(OH)2 are
= Mol HBr initial - mol HBr that reacted with NaOH
= 0.126 mol - 0.039 mol
= 0.087 mol

Zn(OH)2 reacted with HBr according this reaction to give ZnBr2 and H2O, remember to find the mol of Zn(OH)2 you only need to compare the reaction coefficient since the coefficient of HBr and Zn(OH)2  are 2:1  thus the moles of Zn(OH)2.

Zn(OH)2 + 2 HBr -> ZnBr2 + 2 H2O

from the reaction we can find moles of Zn(OH)2
= 1/2 x moles of HBr
= 1/2 x 0.087 mol
= 0.0435 mol

and the mass
= mol x mr
= 0.0435 mol x 99.396
= 4.324 g

Entry Filed under: Titration