hat is the best way to compare a concentrated solution to a dilute solution, given the same volume of solution?
1 answer:
You might be interested in
Answer:
grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions
Explanation:
We will first write the balanced equation for this scenario
3 CaCl2 + 2 Na3PO4 ----> 6 NaCl + Ca3 (PO4)2
3 Mg(NO3)2 + 2 Na3PO4 -----> 6 NaNO3 + Mg3 (PO4)2
The ratio here for both calcium chloride and magnesium nitrate is
The number of moles of each compound is equal to
Using the mole ratio of 3:2, convert each to moles of sodium phosphate.
mole of CaCl2 is equal to
Na3PO4
mole of CaCl2 is equal to
Na3PO4
Converting moles of sodium phosphate to grams of sodium phosphate we get
g/mol
grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions
Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation
Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.