Question

What volume of 0.100 m naoh is required to precipitate all of the nickel (ii) ions from 150.0 ml of a 0.244 m solution of ni(no3)2?

Answer 1

Ni(NO₃)₂(aq) + 2NaOH(aq) → Ni(OH)₂(s) + 2NaNO₃(aq)

n{(Ni(NO₃)₂}=c{(Ni(NO₃)₂}v{(Ni(NO₃)₂}

n{NaOH}=2n{(Ni(NO₃)₂}=2c{(Ni(NO₃)₂}v{(Ni(NO₃)₂}

n{NaOH}=c{NaOH}v{NaOH}=2c{(Ni(NO₃)₂}v{(Ni(NO₃)₂}

v{NaOH}=2c{(Ni(NO₃)₂}v{(Ni(NO₃)₂}/c{NaOH}

v{NaOH}=2*0.244mmol/mL*150.0mL/0.100 mmol/mL= 732 mL