when 15.12 ml of 0.1194 M NaOH is added to 25.00 ml of 0.1246 M HX ( an unknown weak acid), the resulting pH is 5.9. What is the Ka of the unknown acid?
............NaOH + HX ==> NaX + H2O
begin.....M x L...M x L.....0....0
end..see below.
NaOH = moles begin.
HX = moles begin.
Reaction produces the smaller moles NaX (which will be the moles NaOH)
The difference larger-smaller = moles HX at end so NaOH will be zero at end and HX will be the moles HX-moles NaOH.
Ka = (H^+)(X^-)/(HX)
Substitute and solve for Ka.
To find the Ka of the unknown acid, we first need to find the concentration of the acid after the reaction between NaOH and HX has taken place. Then, we can use the equation for the dissociation of the acid to calculate the Ka.
Step 1: Calculate the moles of NaOH and HX
Moles of NaOH = volume (in L) x molarity = 0.01512 L x 0.1194 mol/L
Moles of HX = volume (in L) x molarity = 0.02500 L x 0.1246 mol/L
Step 2: Determine the limiting reagent
Compare the moles of HX and NaOH. The reactant with fewer moles is the limiting reagent.
In this case, moles of HX < moles of NaOH, so HX is the limiting reagent.
Step 3: Calculate the moles of HX remaining
Moles of HX remaining = initial moles of HX - moles of NaOH used
Step 4: Calculate the concentration of HX remaining
Concentration of HX = moles of HX remaining / volume (in L) = moles of HX remaining / 0.02500 L
Step 5: Calculate the concentration of the conjugate base
Since NaOH is a strong base, it will completely dissociate into Na+ and OH-. The moles of OH- will be equal to the moles of NaOH used, and its concentration will be the same as the concentration of NaOH.
Step 6: Set up the dissociation equation
HX (aq) ⇌ H+ (aq) + X- (aq)
Step 7: Write the expression for the Ka
Ka = [H+][X-] / [HX]
Step 8: Calculate the concentration of H+ and X-
[H+] = concentration of HX remaining
[X-] = concentration of OH-, which is the same as the concentration of NaOH used
Step 9: Calculate the Ka
Ka = ([H+][X-]) / [HX]
Now, substitute the values into the equation and calculate the Ka.