" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
[H⁺] = 10^(-pH) = 10⁻² ⁸⁷ = 0,00135 mol/dm3
0,00135mola kwasu ------- 1dm3
x mola kwasu -------------- 5,55dm3
x = 0,00749mola kwasu
Masa NaOH = 40g/mol
1 mol NaOH ----- 40g
x mola NaOH ------- 0,150g
x = 0,00375mola NaOH
0,00749mola - 0,00375mola = 0,00374mola
całkowita objętość roztworu to 5,55dm3
liczymy stężenie molowe
0,00374mola -------- 5,55dm3
x mola -------------------- 1dm3
x = 0,00067mola/dm3
mamy nadmiar kwasu, liczymy pH
pH = -log[H⁺]
pH = -log(0,00067)
pH = 3,17