Answer
m NaOH = 1.4 g
Work Step by Step
First, we need to calculate the number of moles of $Cd^{2+}$.
$n_{Cd^{2+}}$=$n_{Cd(NO_{3})_{2}}$= $C*V$=$0.500M*0.035L$=$0.0175mol$
Reaction: $Cd(NO_{3})_{2}(aq)+ 2NaOH(aq) -> Cd(OH)_{2}(s)+ 2NaNO_{3}(aq)$
This means the reaction ratio is: $n_{NaOH}:n_{Cd(NO_{3})_{2}}=2:1$
$n_{NaOH}=2*0.0175 mol=0.035 (mol)$
$m_{NaOH}=40 g/mol*0.035 mol=1.4 (g)$
