Prepare Solution A (braceros green indicator in acetic acid): Pipet into a 250 m

Question

Prepare Solution A (braceros green indicator in acetic acid): Pipet into a 250 mL volumetric flask the following: 10.0 mL of 3 times 10^-4 M broccoli green solution 25.0 ml of 1.60 M acetic acid (HAc) 10.0 mL of 0 200 M KCl solution Place 5.0 mL of solution A into the beaker containing solution B. (Use a pipet) When solution A is added to Solution 8 the resulting solution is an acetic acetate buffer solution This can be expressed as: We can also derive and equation to give the ratio. N_ac = moles acetate present in solution B mole A_c Molarity HAc in Solution A

Explanation / Answer

I am showing the calculation for one row, thw second one which corresponds to the addition of 10mL of solution A.

Conc. of acetic acid in solution A = 0.16 M

So, moles of acetic acid present, NHAc= 0.16*0.01 = 16*10-4 moles

Conc. of sodium acetate in solution B = 0.0064 M

So, moles of sodium acetate in this solution, NAc = 0.0064*0.25 = 16*10-4 moles

So, NHAc/NAc = 1

So, log (NHAc/NAc) = 0

Thus, pKHAc = pH = 5.12

Now, [HIn]/[In-] = (1.22-0.43)/(0.43-0.01) = 1.88

So, log ([HIn]/[In-]) = 0.274

So, pKHIn = pH + 0.274 = 5.12 + 0.274 = 5.394

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---