Answer:
pH = 7.51
Step-by-step 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]
pH = 7.51