Diketahui terdapat 100mL larutan yang mengandung CH3COOH 0,01mol dan CH3COONa 0,02mol dan Ka = 10^-5. tentukan:
a. ph larutan
b. ph larutan bila ditambah 0,001mol HCl
c. ph larutan bila ditambah 0,001mol NaOH
a. ph larutan
b. ph larutan bila ditambah 0,001mol HCl
c. ph larutan bila ditambah 0,001mol NaOH
abdulqarriem
PH = pKa + log ([CH3COO^-]/[CH3COOH]) pH = 5 + log (0.1/0.1) pH = 5 setelah penambahan HCl, CH3COO^- + H^+ → CH3COOH mol CH3COOH = 900 * 0.1 + 100 * 0.1 = 100 mmol CH3COO^- = 900 * 0.1 - 100 * 0.1 = 80 mmol pH = pKa + log ([CH3COO^-]/[CH3COOH]) pH = 5 + log (80/100) pH = 4.9
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abdulqarriem
(d).
CH3COOH + OH^- → CH3COO^- + H2O
mol CH3COOH = 100 * 0.01 - 0.001 = 0.999 mmol
CH3COO^- = 100 * 0.02 + 0.001 = 2.001 mmol
pH = pKa + log ([CH3COO^-]/[CH3COOH])
pH = 5 + log ([2.001]/[0.999])
pH = 5.3
yahyas11
gak paham nih giman cara ngerjain nya