U 30 m3 vode dodato je 2 tone (2 000 kg) HCl (33 %), koliko je potebno dodati NaOH da bi rastvor imao pH=7?
NaOH + HCl → NaCl + H2O
m(HCl) = w(HCl,ot.)*m(ot.) = 0,33*2 000 000g = 660 000g
Dodatkom vode se masa HCl ne mijenja.
m(NaOH) = m/M(HCl)*M(NaOH) = 660 000g/36,45g mol-1*40 g mol-1 = 724 279,84g
powered by Drupal
Odgovor
NaOH + HCl → NaCl + H2O
m(HCl) = w(HCl,ot.)*m(ot.) = 0,33*2 000 000g = 660 000g
Dodatkom vode se masa HCl ne mijenja.
m(NaOH) = m/M(HCl)*M(NaOH) = 660 000g/36,45g mol-1*40 g mol-1 = 724 279,84g