1.5mg=0.0015g1.5mg=0.0015g
n(NaOH)=0.0015/40=3.75*10^(-5)moln(NaOH)=0.001540=3.75⋅10−5mol
c_1v_1=c_2v_2c1v1=c2v2, where cc is concentration and vv is volume
25mL=0.025L=0.025dm^325mL=0.025L=0.025dm3
[NaOH]=(3.75*10^(-5))/0.025=0.15molcolor(white)(l)^dm(-3)[NaOH]=3.75⋅10−50.025=0.15molldm(−3)
c_2=(c_1v_1)/v_2=(0.15*0.025)/0.250=0.015molcolor(white)(l)dm^(-3)c2=c1v1v2=0.15⋅0.0250.250=0.015molldm−3
(c_av_a)/(n_a)=(c_bv_b)/(n_b)cavana=cbvbnb
As the mole ratio of NaOH:HCl is 1:1, n_a=n_b=1na=nb=1
c_av_a=c_bv_bcava=cbvb
c_a=(c_bv_b)/(v_a)=(0.025(0.015))/(0.020)=0.01875molcolor(white)(l)dm^(-3)ca=cbvbva=0.025(0.015)0.020=0.01875molldm−3