Question #28b04

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Answer 1 · 2017-12-13T15:17:54

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

#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)#