Aqueous zinh chloride (ZnCl2) reacts with aqueous sodium hydroxide (NaOH) lớn produce zinh hydroxide ( Zn(OH)2 ) và sodium chloride (NaCl). Zinc hydroxide is a trắng colour precipitate and it is also an amphoteric hydroxide. Therefore, zinc hydroxide is dissolved when excess NaOH is added lớn the precipitate.

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ZnCl2 + NaOH = Zn(OH)2 + NaCl

When both solutions are mixed, a White colour precipitate, zinc hydroxide ( Zn(OH)2 ) is formed. In aqueous phase, there are sodium ions, chloride ions, little bit of Zn2+ ions & OH- ions.

Balanced equation of ZnCl2 + NaOH reaction

ZnCl2 + 2NaOH = Zn(OH)2 + 2NaCl

From the balanced equation, we can say, 1 mol of zinch chloride reacts with 2 mol of sodium hydroxide and produce 1 mol of zinh hydroxide và 2 mol of sodium chloride.

How to lớn balance the ZnCl2 và NaOH reaction?

This equation can be balanced easily from following steps.

ZnCl2 + NaOH = Zn(OH)2 + NaCl

There are two chlorine atoms in the left side. Therefore, to lớn make two chlorine atoms in the right side, make two NaCl as below. ZnCl2 + NaOH = Zn(OH)2 + 2
NaCl Now, there are two sodium atoms in the right side. To make two sodium in the left side, make two NaOH. ZnCl2 + 2NaOH = Zn(OH)2 + 2NaCl

ZnCl2 & excess NaOH

Remember that zinc is an amphoteric element và zinc hydroxide is an amphoteric hydroxide.

When aqueous NaOH is added khổng lồ ZnCl2 solution drop by drop, at one time, Zn(OH)2 precipitate is formed. If you add more aqueous NaOH to lớn the solution which contains the precipitate, precipitate is dissolved & give sầu a colourless solution due to formation of sodium zincate (Na2ZnO2). Sodium zincate is a colourless aqueous solution.

Zn(OH)2 + 2NaOH → Na2ZnO2 + 2H2O

Sometime above sầu equation is written as below.

Zn(OH)2 + 2NaOH → Na2

After dissolving Zn(OH)2 precipitate in the excess NaOH, what will happen if I add dilute HCl acid?

If you slowly add HCl acid khổng lồ the 2+, you can see a white precipitate is formed in the solution. That is zinc hydroxide precipitate. If you add more HCl acid, that trắng precipitate is also dissolved and give colourless ZnCl2 solution.

What will happen if little amount of ZnCl2 solution is added to lớn large volume of aqueous NaOH?

Due to presence of excess NaOH, reaction will kết thúc up with soluble sodium zincate.

Questions asked by students

Ask your question and find the answer miễn phí.

How bởi I decide whether a precipitate is formed or not when NaCl & ZnCl2 solutions are mixed with each other?

Option 1: If you know the concentratins of each solution, you can vì chưng a calculation to lớn check Ksp value related to the Zn(OH)2.

In the final solution, if * 2 exceeds Ksp of Zn(OH)2, a precipiates is formed. When are taken, you should assume that there is no precipiate và all Zn2+ and Cl- ions exist in aqueous solution.

In the laboratory, I want lớn try this experiment. But, I could not find ZnCl2 in the lab. However, there was ZnO & other comtháng chemicals. How bởi vì I bởi vì the experiment from ZnO?

You can do the experiment by starting from ZnO. Find dilute HCl bottle. Remember that it is good lớn use solid ZnO to prsự kiện any impurities. Measure the weight ZnO and put it in a beaker. Then slowly add dilute HCl solution until all ZnO is dissolved. When all ZnO is dissolveed, ZnCl2 solution is given as the following reaction.

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ZnO + 2HCl → ZnCl2 + H2O

Now, you can continue the rest of the experiment with by adding NaOH lớn ZnCl2 solution.

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