The balanced equation for the reaction between NaOH and HCl is as follows
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
number of NaOH moles reacted - 0.1 mol/L x 0.054 L = 0.0054 mol
number of HCl moles reacted = number of NaOH moles reacted
since the molar ratio of acid to base is 1:1
therefore number of HCl moles reacted - 0.0054 mol
the number of moles of HCl in 125 mL - 0.0054 mol
therefore number of HCl moles in 1000 mL - 0.0054 mol / 125 mL x 1000 mL
number of moles of solute in 1000 mL is known as the molarity concentration
concentration of HCl is 0.0432 M
The symbol is ba. The number of neutrons equal the number of protons
The value of current solubility product is calculated as below
K = (Ag+)( Cl-)
Ag+ = 0.01 M
Cl-= 1 x10^-5M
K is therefore = 1 x10^-5 x 0.01 = 1 x10 ^ -7 M
The K obtained is greater than Ksp
that is K> KSp
1x10^-7 > 1.7 x10 ^-10
will precipitation of AgCl form?
yes the precipitation of AgCl will be formed since K> KSP
