Ćwiczenie nr 12 |
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WYZNACZANIE STAŁEJ DYSOCJACJI WSKAŹNIKÓW ALKACYMETRYCZNYCH METODĄ SPEKTROFOTOMETRYCZNĄ |
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wersja dla wskaźnika dwubarwnego; wszystkie roztwory winny mieć to samo stężenie całkowite wskaźnika |
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Zakładam, że widma są dobrze wyznaczone, a dla pH=2 i pH=12 odpowiadają czystym HA i A- |
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Zatem znam macierz S. |
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[H+] wyznaczam na podstawie pomiaru pH (jeśli nie mam zaufania do pH buforu) |
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wskaźnik: |
wypełnić |
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zakres pH: |
wypełnić |
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Vdod [ml]= |
wypełnić |
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Vcałk. r-ru [ml]= |
wypełnić |
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bufor: |
wypełnić |
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pH \ dł.fali |
400 |
420 |
440 |
460 |
480 |
500 |
520 |
540 |
560 |
580 |
600 |
620 |
640 |
660 |
680 |
700 |
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poniżej wpisujemy widmo wskaźnika w postaci niezdysocjowanej (pH = ok. 2) |
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0 |
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macierz A |
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poniżej wpisujemy widma wskaźnika dla pH z zakresu zmiany barwy |
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1 |
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2 |
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3 |
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4 |
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poniżej wpisujemy widmo wskaźnika w postaci zdysocjowanej (pH = ok. 12) |
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5 |
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1) macierz S (pH=2 i pH=12) |
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0 |
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2) macierz ST |
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1 |
0 |
0 |
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3) macierz stężeń C = A.ST.(S.ST)-1 |
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2 |
0 |
0 |
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1-a |
a |
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3 |
0 |
0 |
nr r-ru |
[HA]/c0 |
[A-]/c0 |
[HA]/c0 + [A-]/c0 |
pH dośw. |
a śr. |
pK |
a śr. obl. z war.([HA]+[A-])/c0=1 |
pK |
UWAGA: |
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4 |
0 |
0 |
1 |
#VALUE! |
#VALUE! |
#VALUE! |
0 |
#VALUE! |
tu wpisać |
#VALUE! |
tu wpisać |
jeśli a < 0, pominąć ten pkt przy obl. średniej pK |
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5 |
0 |
0 |
2 |
#VALUE! |
#VALUE! |
#VALUE! |
0 |
#VALUE! |
odpowiedni |
#VALUE! |
odpowiedni |
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6 |
0 |
0 |
3 |
#VALUE! |
#VALUE! |
#VALUE! |
0 |
#VALUE! |
wzór |
#VALUE! |
wzór |
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7 |
0 |
0 |
4 |
#VALUE! |
#VALUE! |
#VALUE! |
0 |
#VALUE! |
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#VALUE! |
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8 |
0 |
0 |
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Im bardziej różne od 1, tym mniej dokł. sporządzone roztwory. |
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średnia = |
#DIV/0! |
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#DIV/0! |
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9 |
0 |
0 |
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odch.std.= |
#DIV/0! |
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#DIV/0! |
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10 |
0 |
0 |
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liczba pomiarów = |
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wpisać prawidłową liczbę |
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11 |
0 |
0 |
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odch.std.średniej= |
#DIV/0! |
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#DIV/0! |
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12 |
0 |
0 |
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poziom ufności= |
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wpisać odpowiedni poziom |
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13 |
0 |
0 |
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liczba stopni swobody= |
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wpisać prawidłową liczbę |
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14 |
0 |
0 |
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t Stud.= |
Err:502 |
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Err:502 |
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15 |
0 |
0 |
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przedział ufności= |
#DIV/0! |
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#DIV/0! |
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16 |
0 |
0 |
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Ka,apr= |
#DIV/0! |
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#DIV/0! |
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