Vs= 25ml=0,025 l
c=0,5mol/l
c=n/V---> n=0,5x0,025=0,0125 mol HCl
acest numar de moli se introduc in borcanul de 0,5l
0,5l.........0,0125molHCl....
1 l.............0,025 molHCl
[HCl]=[H⁺]= 2,5x10⁻²mol/l
pH= -log[H⁺]= 2-log2,5=.2-0,4=1,6-este pH-ul solutiei care se adauga, netinand cont de pH-ul conservei
1mol..............1mol
HCl + NaHCO3---> NaCl + H2CO3
0,025mol.....x=0,025mol sare
m=nxM= 0,025(23+1+12+3X16)g=...................calculeaza !!!!
