Ask a New Question

Asked by Ai

You have prepared a 1.50 M ZnCl2 solution. The pH must be below what value to avoid precipitation of zinc hydroxide in the solution? The solubility product of Zn(OH)2 is 3.0 x 10−16.
3 years ago

Answers

Answered by DrBob222
Zn(OH)2 ----> Zn^2+ + 2OH^-
Ksp = 3E-16 = (Zn^2+)(OH^-)^2
Plug in 1.5M = (Zn^2+) and solve for (OH^-) and convert that to pH. Post your work if you get stuck.
3 years ago

Related Questions

When NH3 is prepared from 28 g of N2 and excess H2, the theoretical yield of NH3 is 34 g. When this... Zn(s) + 2 HCl(aq) H2(g) + ZnCl2(aq) When 25.0 g of Zn reacts, how many L of H2 gas are formed at S... Fe can be prepared as 2 Al + Fe2O3 ----> 2 Fe + Al2O3. Suppose that 0.450 moles of Fe2O3 are reacted... You have prepared a 400 mL of a .210 M acetate buffer solution with a pH of 4.44. 1. Determine the... You have prepared 20.00 mL of a buffer, using 4.00 mL 0.010 M HA (weak acid) and 3.00 mL 0.010 M NaA... Now that you have prepared for your meeting, you are required to cor a session chairing a technical... 2HCL(aq) + Zn(s) -> ZnCl^2(aq) + H^2(g) the molar mass of hydrogen is 1 g/mol, the molar mass of ch... What should you come prepared to discuss during a doctor's appointment? frequency of symptoms seve... is Zn+2HCL--->ZnCl subscript2+H subscript2
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use