The amount of millimoles of solid naoh that must be added to 50.0 milliliters is: 5.0 mmol of NaOH.
The [H+] at pH 7.49 is:
[H+] = 10–^7.49
[H+]= 3.24×10–^8 M
And when the solution is half-neutralized pH will be:
pH = pKa while [OCl– ]÷[HOCl] = 1
Now let find the millimoles of solid naoh:
0.20 mol HOCl÷1 L× 50.0 mL
= 10.0 mmol HOCl
Half of 10.0 mmol HOCl will be added
Millimoles of solid naoh=10.0 mmol HOCl÷2
Millimoles of solid naoh= 5.0 mmol of NaOH
Inconclusion the amount of millimoles of solid naoh that must be added to 50.0 milliliters is: 5.0 mmol of NaOH.
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