Science hw question?
Explain how you would prepare a buffer with a pH of 4.80 using CH3COOH (1.8x10^-5) CH3COONa system?
please explain, I do not understand. This is a chemistry problem btw
Explain how you would prepare a buffer with a pH of 4.80 using CH3COOH (1.8x10^-5) CH3COONa system?
please explain, I do not understand. This is a chemistry problem btw
kumorifox
You use the Henderson-Hasselbalch equation for this.
pH = pKa + log ([A^-]/[HA]), therefore [A^-]/[HA] = 10^(pH - pKa)
The Ka of acetic acid is 1.8×10^-5, therefore its pKa is the negative log of that, or 4.74. Thus, in this case, [A^-]/[HA] = 10^(4.80-4.74) = 10^0.06 = 1.15.
You need a ratio of 1:1.15 for acetic acid:sodium acetate. You can achieve this by mixing, say, equal volumes of 0.1 M acetic acid and 0.115 M sodium acetate. There are many more possibilities that will give the same pH buffer, but the buffer capacity will vary.