Answer:

Explanation:
Hello,
In this case, for the dissociation of calcium fluoride:

The equilibrium expression is:
![Ksp=[Ca^{2+}][F^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BCa%5E%7B2%2B%7D%5D%5BF%5E-%5D%5E2)
In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent
is computed as follows:

Thus, the molar solubility equals the reaction extent
, therefore:

Regards.
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.
C
a mountain range will form
Answer:
your answer gonna be 20 miles