Change in pH when NaBrO is added to solution?

I need to calculate the change in pH when 6.80 g of sodium hypobromite (NaBro?) is added to 100 mL of 0.50 M HBrO (aq) (which has 4.50 pH).



It tells me to ignore any change in volume.



How do I go about this?Change in pH when NaBrO is added to solution?
First, figure the pKa of HOBr. Use

[H+] = 鈭歔Ka*c]

(this can derived from the equilibrium equation)

Since pH = -log[H+],

10^(-4.50) = 鈭歔Ka*(0.50)];

solving, Ka = 2.00x10^-9, or pKa = 8.76 for HOBr.

The molecular mass of NaOBr is 118.9 g/mole, so the concentration when 6.8 g is dissolved in 100 mL is:

(6.8 g) / (118.9 g/mole) (0.1 L) = 0.57 M.

Now use the Henderson - Hasselbalch equation:

pH = pka + log ([NaOBr]/[HOBr])

pH = 8.76 + log(0.57/0.50) = 8.76.

The change in pH is:

8.76 - 4.50 = 4.26.

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