zwracam sie z prosba o spr zadań
1. oblicz pH :
A] Ba(OH)2 (\(\displaystyle{ \alpha = 100}\)% ) o C= 2*\(\displaystyle{ 10^{-4}}\) mol/dm3
K=\(\displaystyle{ \alpha ^{2}}\) * C
K=2*\(\displaystyle{ 10^{-4}}\)
\(\displaystyle{ [H+] ^{2}}\)= K * C
B] HNO3 o C= 0,005 mol/dm3
[H+]= 0,005
Ph= - log(0,005) > mam problem z obliczeniem tego wyniku
c] amoniaku o C = 0,001 mol/dm3 (K= 1,75 * \(\displaystyle{ 10^{-5}}\))
\(\displaystyle{ [H+] ^{2}}\)= K * C
\(\displaystyle{ [H+] ^{2}}\)= 1,75 * \(\displaystyle{ 10^{-5}}\) * \(\displaystyle{ 10^{-3}}\)
mam problem z obliczeniami
pH= - LOG [H+]
2.oblicz pH 0,5 MOLOWEGO CH3COOH , ktorego K=1,8* \(\displaystyle{ 10^{-5}}\)
\(\displaystyle{ [H+] ^{2}}\)= K * C
\(\displaystyle{ [H+] ^{2}}\)= 0,9* \(\displaystyle{ 10^{-5}}\)
[H+]=0.003
pH= - log (0.003) tu problem z obliczeniem