4HW 68% 02/27/18 You have a 1.153 g sample of an unknown solid acid, HA,dissolve

Question

4HW 68% 02/27/18 You have a 1.153 g sample of an unknown solid acid, HA,dissolved in enough water to make 20.00 mL of solution. HA reacts with KOHaa) according to the following balanced chemical equation: HA(aq) + KOH(aq)-, KA(aq) + H2O(1) 1st attempt Part 1 (2 points) See Periodic Table ’ See Hint If 15.15 mL of 0.440 M KOH is required to titrate the unknown acid to the equivalence point, what is the concentration of the unknown acid? ? See Hint What is the molar mass of HA? g/mol K28/37 30F 37

Explanation / Answer

1)
Balanced chemical equation is:
KOH + HA ---> KA + H2O


Here:
M(KOH)=0.44 M
V(KOH)=15.15 mL
V(HA)=20.0 mL

According to balanced reaction:
1*number of mol of KOH =1*number of mol of HA
1*M(KOH)*V(KOH) =1*M(HA)*V(HA)
1*0.44*15.15 = 1*M(HA)*20.0
M(HA) = 0.3333 M

Answer: 0.3333 M

2)
volume , V = 20.00 mL
= 2*10^-2 L


use:
number of mol,
n = Molarity * Volume
= 0.3333*0.02
= 6.666*10^-3 mol

mass(solute)= 1.153 g


use:
number of mol = mass / molar mass
6.666*10^-3 mol = (1.153 g)/molar mass
molar mass = 173 g/mol
Answer: 173 g/mol

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