0.5 mol of HA (Ka = 1.5×10-3) was added to 3.0 L of water. What is the pH? After?
0.5 mol of HA (Ka = 1.5×10-3) was added to 3.0 L of water. What is the pH? After
equilibrium was reached, 1.0 g of NaOH was added to the solution. What is the new [H+]?
I did the first part and I got ph = 1.82...which was right..But I'm struggling with the second part...help
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Answers & Comments
Convert the mass of sodium hydroxide to moles (.025mol) and do an ICE table (initial change end) to determine how much HA moles will be left over (.5-.025), then calculate the [H+] using the new mol data instead of the .5
1.0g of NaOH in 3 L of water has a concentration of:
moles of NaOH = 3 / 40 = 0.075
Conc of NaOH is 0.075 moles / 3 L = 0.025 mole per L
These two substances react in a 1 to 1 ratio, So for ever 0.025 mole of NaOH added you will use up 0.025 mole of H+ ions.