Q: A student standardizes sodium hydroxide by weighing out 1.207 grams of potassium hydrogen iodate, dissolving the compound in water and titrating to a suitable endpoint:
If the volume to reach the endpoint is 32.80 mL, what is the concentration of sodium hydroxide?
The molar mass of PHI is $389.75 \text{g/mol}$. if 1.207g is added to the titrand beaker, then there are: $$\frac{1.207\text{g}}{389.75\text{g/mol}} = 0.0031\text{mol of KHI in the titrand beaker}$$
NaOH reacts with KHI in a 1:1 molar ratio, and 32.8mL of NaOH was added to reach the endpoint. This means that 0.0031mol of NaOH was added. So, the concentration of NaOH is:
$$\frac{0.0031\text{mol}}{0.0328\text{L}} = 0.0945\text{M}$$
This is coming up as a wrong answer in my book. Can anyone explain where I'm going wrong?
