Calculate the pH of the solution that results from each of the following mixtures. PART A 150.0 mL of 0.26 M HF with 230.0 mL of 0.32 M NaF The Ka of hydrofluoric acid is 6.8 x 10−4. Express your answer using two decimal places.
PART B 170.0 mL of 0.11 M C2H5NH2 with 270.0 mL of 0.22 M C2H5NH3Cl. Express your answer using two decimal places.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

4d^55s^1 ( third choice)