Answer:
pH = 12.33
Explanation:
Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.
The titration reaction is
HA + KOH ---------------------------- A⁻ + H₂O + K⁺
number of moles of HA : 118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA
number of moles of OH : 115.4 mL/1000ml/L x 0.400 mol/L = 0.046 mol A⁻
therefore the weak acid will be completely consumed and what we have is the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.
n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH
pOH = - log (KOH)
M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M
pOH = - log (0.0021) = 1.66
pH = 14 - 1.96 = 12.33
Note: It is a mistake to ask for the pH of the <u>acid solutio</u>n since as the above calculation shows we have a basic solution the moment all the acid has been consumed.
I’m pretty sure it’s true
K gives 1 electron, and Br takes 1 electron.
<span> aluminum is an element. All elements are pure substances, so that means they are homogenous.
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</span>
Actual yield of Fe2(So4)3 = 18.5g
2FePo4 + 3Na2SO4 -> Fe2(SO4)3 + 2Na3PO4
Mole of FePO4 = mass of it / its molar mass =
25 g / (55.8 + 31 + 16*4) = 0.166 mol
every 2 mole of FePO4 will form 1 mole of Fe2(SO4)3
Mole of Fe2(SO4)3 produced = 0.166 / 2 = 0.0829 mol
0.0829 * (55.8*2 + 3*(32.1+ 16*4)) = 33.148 g of Fe2(SO4)3
18.5 / 33.148 * 100 = 55.8%
