Asked by BILL
When 100mL of 0.60mol per liter K3BO3 is added to 200 mL of 0.40 mol per liter Ca3(BO3)2 Find
a)moles of K3BO
b) moles of Ca(BO3)2
c)final concentration of K+
d)final concentration of Ca+2
e) final concentration of BO3 negative 3
a)moles of K3BO
b) moles of Ca(BO3)2
c)final concentration of K+
d)final concentration of Ca+2
e) final concentration of BO3 negative 3
Answers
Answered by
DrBob222
I'm assuming Ca3(BO3)2 has no Ksp. I couldn't find one.
a. mols = M x L = ?
b. mols = M x L = ?
c. mols K = 3 x mols K3BO3 and (K^+) = mols K/total L
d. mol BO3^3- from K3BO3 = mols K3BO3
mols BO3^3- from Ca3(BO3)2 = 2 x mols Ca3(NO3)2.
Then (BO3^3-) = total mols BO3^3-/total L
a. mols = M x L = ?
b. mols = M x L = ?
c. mols K = 3 x mols K3BO3 and (K^+) = mols K/total L
d. mol BO3^3- from K3BO3 = mols K3BO3
mols BO3^3- from Ca3(BO3)2 = 2 x mols Ca3(NO3)2.
Then (BO3^3-) = total mols BO3^3-/total L