The pH of buffer solutions. Weight of NaC2H3O2*3H2O =3.47g pH of original buffer
ID: 891193 • Letter: T
Question
The pH of buffer solutions.
Weight of NaC2H3O2*3H2O =3.47g
pH of original buffer soution = 4.53
pH of buffer + HCl = 3.94
pH of buffer + NaOH = 5.00
Additional details: approximately 8.8 mL of 3M acetic acitd was added to the 3.47g of NaC2H3O2*3H2O. And approximately 56.0mL of distilled water. (This makes the buffer solution) This solution was then split in half. To one half 1 mL of 3M HCl was added. To the second half 1mL of 3M Na OH was added.
Calculate the pH of buffer.
Calculate the pH of buffer + HCl
Calculate the pH of buffer + NaOH
Please show all calculations and work.
please help me. I confused
Explanation / Answer
Mass of NaC2H3O2*3H2O = 3.47 g
Molar mass of NaC2H3O2*3H2O = 136.07 g/mol
Moles of NaC2H3O2*3H2O = 3.47 g / 136.07 g/mol
= 0.0255 moles
Molarity of acitic acid = 3 M
Volume of acetic acid = 8.8 mL = 0.0088 L
Moles of acetic acid = 3.0 * 0.0088
= 0.0264 moles
Total volume of buffer = 56.0 + 8.8
= 64.8 mL = 0.0648 L
[HA] = 0.0264 / 0.0648
= 0.407 M
[A-] = 0.0255 / 0.0648
= 0.393 M
pKa of acetic acid = 4.76
pH = pKa + log [A-] / [HA]
= 4.76 + log(0.393 / 0.407)
= 4.76 - 0.0152
= 4.744
When this solution is divided in two halves.
Volume of each half = 32.4 mL = 0.0324 L
Volume of HCl added = 1 mL = 0.001 L
Molarity of HCl = 3 M
Moles of HCl = 3 * 0.001
= 0.003 M
Total moles of H+ = 0.0264 + 0.003
= 0.0294 moles
Total volume = 32.4 + 1.0 = 33.4 mL
= 0.0334 L
[A-] = 0.0255 / 0.0334
= 0.763 M
[HA] = 0.0294 / 0.0334
= 0.880 M
pH = 4.76 + log (0.763 / 0.880)
= 4.76 - 0.062
= 4.698
When 1mL of 3M Na OH was added to other half.
Moles of OH- = 0.003
Total moles of A- = 0.0255 + 0.003
= 0.0285
[A-] = 0.0285 / 0.0334
= 0.853 M
[HA] = 0.0264 / 0.0334
= 0.790 M
pH = 4.76 + log (0.853 / 0.790)
= 4.76 + 0.033
= 4.76
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