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Calculate the pH after adding 5.00 mL of 0.15 M HNO3 to 250 mL of an...

Calculate the pH after adding 5.00 mL of 0.15 M HNO₃ to 250 mL of an acetic acid/acetate buffer with an initial pH of 4.32 and a total buffer concentration of 0.25 M. Neglect activity coefficients for this problem.

Expert Solution

Moles of HNO3 added =0.15 mol/L *5mL/1000mL = 0.00075 moles

Calculate the moles of acetate and acetic acid initially present. pka of acetate buffer = 4.76 (refer to your textbook)

pH = pka + log [salt/acid]

4.32 = 4.76 + log [salt/acid]

[salt/acid] = 0.36

decimal fraction of salt = 0.36/ 1+ 0.36 = 0.26

decimal fraction of acid = 1/1+ 0.36 = 0.74

Molarity of acetate =0.26 *0.25 = 0.065 M

Number of moles of acetate = 0.065 M * 250mL/1000mL = 0.016 moles

Molarity of acetic acid = 0.74 * 0.25 M

Number of moles of acetic acid = 0.74 * 0.25 M * 250mL/1000mL = 0.046 moles

HNO3 added will react with acetate to convert it to acetic acid. So, you have to substract the number of moles of acetate and add it to number of moles of acetic acid.

Moles of acetate after adding HNO3 = 0.016 moles - 0.00075 moles = 0.015 moles

Molarity of acetate in after adding HNO3 = 0.015 moles *1000ml/(5 + 250mL) = 0.059 M

Moles of acetic acid afetr adding HNO3 = 0.046 + 0.00075 =0.047 moles

Molarity of acetic acid after adding HNO3 = 0.047 /0.255 L = 0.18 M

pH = pKa + log [salt/acid]

= 4.76 + log [0.059/0.18] = 4.28

ph of the medium will be 4.28.

queen_honey_blossom answered 1 year ago

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