A buffer is made by mixing 2.0 grams of sodium acetate (1mol/82g) and 10mL of 3.
ID: 496396 • Letter: A
Question
A buffer is made by mixing 2.0 grams of sodium acetate (1mol/82g) and 10mL of 3.0 M acetic acid. Then 50 mL of water is added to the mixture. Find the pH of this buffer. (2 methods ICE table and henderson hasselbach equation)
Now, let's say that I'd like to split this 60mL of buffer into two portions (30mL each)
1. one half of the solution will be added with 1mL of HCl (6.0 M)
2. The other half of the solution will be added with 1mL of NaOH (6.0 M)
Find the pH for solution 1 and 2. Please show steps
Explanation / Answer
moles of sodium acetate = 2 / 82 = 0.02439
concentration of sodium acetate = 0.02439 / 0.06 = 0.4065 M
moles of acetic acid = 10 x 3 / 60 = 0.5 M
pH = pKa + log [salt / acid]
= 4.74 + log [0.4065 / 0.5]
= 4.65
pH = 4.65
1)
concnetration of HCl = 0.006
pH = pKa + log [salt - C / acid + C]
= 4.74 + log [0.4065 - 0.006 / 0.5 + 0.006]
= 4.64
pH = 4.64
2)
concentration of NaOH = 6 x 1 /1000 = 0.006
pH = pKa + log [salt + C / acid - C]
= 4.74 + log [0.4065 + 0.006 / 0.5 - 0.006]
= 4.66
pH = 4.66
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