Consider the following pH indicators and their pH range: Methyl orange: 3.2-3.4
ID: 991511 • Letter: C
Question
Consider the following pH indicators and their pH range:
Methyl orange: 3.2-3.4
Methyl red: 4.8-6.0
Bromothymol blue: 6.0-7.6
Phenolphthalein: 8.2-10.0
Alizarin yellow: 10.1-12.0
Assume an indicator works best when the equivalence point of a titration comes in the middle of the indicator range. Which indicator is the best choice to use when 0.100 M NH3 (Kb = 1.8 X 10^-5) is titrated with 0.100 M HCl?
The answer is Methyl Red but I am confused as to why that is because:
pOH = pKb + log(acid/base) correct?
So, -log(1.8 X10^-5) = 4.74 + 0 = 4.74 = "POH"!
pOH = 4.74
pH = 14 - 4.74 = 9.26, so Phenolphthalein is what I chose.
It chose Methyl red as if 4.74 is the pH? that's the pOH! How did they go from Kb to pH without converting. I am very confused, please advise!
Explanation / Answer
Henderson-Hasselbalch Equation
pH = pKa + log ([A-]/[HA])
[A-] = molar concentration of a conjugate base
[HA] = molar concentration of a undissociated weak acid (M)
The equation can be rewritten to solve for pOH:
pOH = pKb + log ([HB+]/[ B ])
[HB+] = molar concentration of the conjugate base (M)
[ B ] = molar concentration of a weak base (M)
But pKb = -logKb
pOH = pKb + log ([HB+]/[ B ]) becomes
pOH = -logKb + log ([HB+]/[ B ]) = -log(1.8 X10^-5) + log (0.1/0.1) = 4.74 + 0 = 4.74 (log 1 = 0)
pOH = 4.74
pH = 14 - 4.74 = 9.26
Which indicator is the best choice to use when 0.100 M NH3 (Kb = 1.8 X 10^-5) is titrated with 0.100 M HCl
Answer: HCl is strong acid and ammonia is a weak base both are same concentration but HCl is a strong acis hence it completely ionizes into ions given full H+ ions in solution whereas NH3 is a weak base so dissociation is very less compare to HCl, results that solution is somewhat acidic. hence using acidic pH indicator is best choice.
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