Question

Question #28b04

Answer 1

`0.01875molcolor(white)(l)dm^(-3)`

Explanation:

`1.5mg=0.0015g`

`n(NaOH)=0.0015/40=3.75*10^(-5)mol`

`c_1v_1=c_2v_2`, where `c` is concentration and `v` is volume

`25mL=0.025L=0.025dm^3`

`[NaOH]=(3.75*10^(-5))/0.025=0.15molcolor(white)(l)^dm(-3)`

`c_2=(c_1v_1)/v_2=(0.15*0.025)/0.250=0.015molcolor(white)(l)dm^(-3)`

`(c_av_a)/(n_a)=(c_bv_b)/(n_b)`

As the mole ratio of NaOH:HCl is 1:1, `n_a=n_b=1`

`c_av_a=c_bv_b`

`c_a=(c_bv_b)/(v_a)=(0.025(0.015))/(0.020)=0.01875molcolor(white)(l)dm^(-3)`