a.
in vasul 3 = solutie de turnesol, indicator acido-bazic care in mediu acid se coloreaza in rosu (vasul 2) iar in mediu bazic se coloreaza in albastru (vasul 2)
b.
vasul 1 = mediu acid => acid sulfuric (H2SO4)
vasul 2 = mediu bazic => hidroxid de potasiu (KOH)
c.
ms= 250g c%= 15%
c% = mdx100/ms
=> md = msxc%/100 = 250x15/100 = 37,5 g H2SO4
M.H2SO4 = 2x1+32+4x16 = 98 g/mol
M.KOH = 39+16+1 = 56 g/mol
M.K2SO4 = 2x39+32+4x16 = 174 g/mol
M.H2O = 2x1+16 = 18 g/mol
37,5 g md g a g b g
H2SO4 + 2KOH --> K2SO4 + 2H2O
98 2x56 174 2x18
=> md = 37,5x2x56/98 = 42,86 g KOH
din c% => ms = mdx100/c%
= 42,86x100/15 = 285,7 g sol (reprezinta x-ul din problema)
d.
compozitia procentuala = c% sol. finala
sol.finala = m.sare.formata(a) + m.apa.sol.H2SO4 + m.apa.sol.KOH + m.apa.formata(b)
m.apa.sol.H2SO4 = ms-md = 250 - 37,5 = 212,5 g
m.apa.sol.KOH = ms(x) - md = 285,7 - 42,86 = 242,84 g
m.apa.formata = b = 37,5x2x18/98 = 13,78 g
m.sare.formata(a) = 37,5x174/98 = 66,58 g K2SO4
=> ms.final = 66,58+212,5+242,84+13,78 = 535,7 g.
=> c%final = mdx100/ms.final
= 66,58x100/535,7 = 12,43%
