care este ph-ul unei solutii apoase care contine 9,125mg HCl dizolvat in 250ml solutie?

M hCl=36,5, n=m/M=9,125/36,5=0,25. C=n/V=0,25/250=0,001M, pH =-lg[C]=-lg[10^(-3)]=3.

Vom transforma masa de HCl in moli. 1 mol de HCl...............36500mg X moli de HCl.............. 9,125mg X=9,125/36500=0,00025moli de HCl; Vom calcula concentratia molara a solutiei Cm=n/Vs=0,00025/0,25=0,001M=1/1000=10^(-3) M. pH este puterea cu semn schimbat a concentratiei => pH=-(-3)=3. Daca ai facut logaritmii, atunci folosesti formula: pH=-lg(Cm)=-lg(10^(-3))=3lg10=3