Answer:
The mass of calcium sulfate that will precipitate is 6.14 grams
Explanation:
<u>Step 1:</u> Data given
500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−
Ksp for CaSO4 is 2.40*10^−5
<u>Step 2:</u> Calculate moles of Ca^2+
Moles of Ca^2+ = Molarity Ca^2+ * volume
Moles of Ca^2+ = 0.10 * 0.500 L
Moles Ca^2+ = 0.05 moles
<u>Step 3: </u>Calculate moles of SO4^2-
Moles of SO4^2- = 0.10 * 0.500 L
Moles SO4^2- = 0.05 moles
<u>Step 4: </u>Calculate total volume
500.0 mL + 500.0 mL = 1000 mL = 1L
<u>Step 5: </u> Calculate Q
Q = [Ca2+] [SO42-]
[Ca2+]= 0.050 M [O42-]
Qsp = (0.050)(0.050 )=0.0025 >> Ksp
This means precipitation will occur
<u>
Step 6:</u> Calculate molar solubility
Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)
2.40 * 10^-5 = x²
x = √(2.40 * 10^-5)
x = 0.0049 M = Molar solubility
<u>
Step 7:</u> Calculate total CaSO4 dissolved
total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g
<u>Step 8:</u> Calculate initial mass of CaSO4
Since initial moles CaSo4 = 0.050
Initial mass of CaSO4 = 0.050 * 136.14 g/mol
Initial mass of CaSO4 = 6.807 grams
<u>Step 9:</u> Calculate mass precipitate
6.807 - 0.667 = 6.14 grams
The mass of calcium sulfate that will precipitate is 6.14 grams
. In this case, conjugate base will react and neutralize it to produce some amount of HF:
.
, the acidic component will react with it to produce some amount of conjugate base:
. The ratio of HF to NaF in this case is held around the same value for addition of small amounts of strong acids/bases, so pH is kept almost constant, while in neutral water, pH would drastically increase or decrease.