Using a 0.25 M phosphate buffer with a pH of 7.0, you add 0.79 mL of 0.49 M HCl to 42 mL of the buffer. What is the new pH of the solution?
Using a 0.25 M phosphate buffer with a pH of 7.0, you add 0.79 mL of 0.49 M NaOH to 42 mL of the buffer. What is the new pH of the solution?
The molar ratio of the buffer is calculated, using the Henry Hasselbach equation cleared:
n Salt / n Acid = 10 ^ (pH - pKa) = 10 ^ (7 - 7.21) = 0.62
It has:
i) n Salt - 0.62 * n Acid = 0
ii) n Salt + n Acid = M * V = 0.25 M * 0.042 L = 0.0105 mol
System of equations is applied and you have:
n Salt = 0.004 mol
n Acid = 0.006 mol
The added moles of HCl are calculated:
n HCl = 0.49 * 0.00079 = 0.0004 mol
HCl reacts with salt (decreasing it) and forms acid (increasing it):
The new pH is calculated:
pH = pKa + log (n Salt / n Acid) = 7.21 + log (0.004 - 0.0004 / 0.006 + 0.0004) = 6.77
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