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Molar solubility is number of moles of the solute that can be dissolved per liter of solution before the solution becomes saturated.
The molar solubility of lead(ii) chloride with ksp value of 2.4 × 10e4 can be solve as:
Ksp = s2 = 2.4 × 10e4
s2 = 2.4 × 10e4
s = √(2.4 × 10e4)
s = 154.9 mol/L
Answer:
0.375 M
Explanation:
NaOH(aq) + HBr(aq) ------------> NaBr(aq) + H2O(l)
Concetration of acid CA= 0.250M
Concentration of base CB= ????
Volume of acid VA= 30.0mL
Volume of base VB= 20.0mL
Number of moles of acid nA= 1
Number of moles of base nB= 1
CA VA/CB VB= nA/nB
CB= CAVAnB/VB nA
= 0.25× 30×1/20×1= 0.375 M