Nickel(II) nitrate reacts with sodium hydroxide to form aprecipitate, nickel(II)

Question

Nickel(II) nitrate reacts with sodium hydroxide to form aprecipitate, nickel(II) hydroxide. What volume of 0.100 M NaOH isrequired to precipitate all of the nickel(II) ion from 50.00 mL of0.175 M nickel(II) nitrate?                                 PLEASE SHOW ALL YOUR WORK Nickel(II) nitrate reacts with sodium hydroxide to form aprecipitate, nickel(II) hydroxide. What volume of 0.100 M NaOH isrequired to precipitate all of the nickel(II) ion from 50.00 mL of0.175 M nickel(II) nitrate?                                 PLEASE SHOW ALL YOUR WORK

Explanation / Answer

Ni(NO3)2 + 2NaOH -> Ni(OH)2+2NaNO3 c(Ni(NO3)2) = n/v n(Ni(NO3)2) cv = 0.175 x 0.05 =0.00875mol n(NaOH) =2/1 x 0.00875 = 0.0175mol c(NaOH) = n/v 0.1 = 0.0175/v v(NaOH) = 0.175L Hope this helps!

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