A 100.0 mL buffer solution is 0.175 M in HClO and 0.150 M in NaClO. What is the pH after addition of 0.090 g of NaOH?A

pH = 7.51

Explanation:

The pKa of the HClO/NaClO buffer is 7.46. To determine the pH of this buffer we can use H-H equation:

pH = pKa + log [A⁻] / [HA]

pH = 7.46 + log [NaClO] / [HClO]

Where [] is molarity -or moles- of each compound

Initial moles of HClO and NaClO:

HClO: 0.100L * (0.175mol / L) = 0.0175 moles HClO

NaClO: 0.100L * (0.150mol / L) = 0.0150 moles NaClO

Now, HClO reacts with NaOH producing NaClO:

HClO + NaOH → NaClO + H₂O

The moles of NaOH that reacts (Molar mass: 40g/mol) are:

0.090g * (1mol / 40g) = 0.00225 moles NaOH.

That means after the reaction, 0.00225 moles of HClO are consumed and 0.00225 moles of NaClO are produced.

And after the reaction, moles are:

Final moles:

HClO: 0.0175 mol - 0.00225 mol = 0.01525 moles

NaClO: 0.0150 mol + 0.00225 mol = 0.01725 moles

Replaing in H-H equation:

pH = 7.46 + log [0.01725moles] / [0.01525moles]

1. they vary with a higher number because boiling is way hotter than melting