Question

what concentration of Zn^2+ will remain when 200mL of 4.50*10^-3 M Zn(NO3)2 is combined with 200.0mL of .250 M NaOH? Kf of Zn(OH)4 ^2- = 2.0*10^15

Answer 1

The reaction involved here is :

Zn(NO₃)₂ + 2NaOH = Zn(OH)₂ + 2NaNO₃

To find out the conc. of Zn²⁺ ion in the mixed solution firstly we have to find the conc. of Zn²⁺ ion in Zn (NO₃)₂ :

( As we know C = n/v)

= 4.50 * 10^_3 M* 200ml * 1 / 1000ml

= 0.90 * 10^_3 M

Now we calculate conc. of OH^_ ion in NaOH

0.250M * 200ml * 2 / 1000ml

= 1.0 * 10^_1 M

Now its given Kf (formation constant ) of Zn(OH)₂ as per reaction

Zn ²⁺ + OH^_ = Zn(OH)₂

so accordingly Kf = conc of Zn(OH)₂ / conc of Zn²⁺ x conc of OH^_

hence conc of Zn(OH)₂ = 2.0 * 10¹⁵ *0.90 * 10^_3 * 1 * 10^_1

                                  = 1.80 * 10¹¹ M

Now the conc of Zn²⁺ ion in Zn(OH)₂ = 1.80 * 10¹¹ * 1 * 400ml /1000ml

                                                    = 7.20* 10¹⁰ M