Answer :
Answer:
#1. M NaOH = 0.1634 M
#2. M NaOH = 0.1464 M
#3. M NaOH = 0.1501 M
Explanation:
- molarity [=] mol/L
balanced reaction:
- NaOH + KHP → KNaP + H2O
∴ molar mass KHP = 204.22 g/mol
#1:
∴ mass KHP = 0.5240 g
∴ volume NaOH = 15.7 mL = 0.0157 L
⇒ moles NaOH = (0.5240 g KHP)(mol/204.22 g KHP)(mol NaOH/mol KHP)
⇒ moles NaOH = 2.566 E-3 mol
⇒ M NaOH = (2.566 E-3 mol)/(0.0157 L) = 0.1634 M
#2:
∴ mass KHP = 0.5320 g
∴ volume NaOH = 17.8 mL = 0.0178 L
⇒ moles NaOH = 2.605 E-3 mol
⇒ M NaOH = 0.1464 M
#3:
∴ mass KHP = 0.5120 g
∴ volume NaOH = 16.7 mL = 0.0167 L
⇒ moles NaOH = 2.5071 E-3 mol
⇒ M NaOH = 0.1501 M
The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium, a state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.
NaOH + KHP → KNaP + H2O
According to the question, the molar mass is KHP = 204.22 g/mol
- mass KHP = 0.5240 g
volume NaOH = 15.7 mL = 0.0157 L
moles NaOH =[tex]0.5240*204.22[/tex]
moles NaOH = 2.566.
Molarity is NaOH =[tex]= \frac{2.566}{0.0157} =0.1634[/tex]
- mass KHP = 0.5320 g
[tex]volume NaOH = 17.8 mL = 0.0178 Lmoles NaOH = 2.605 molM NaOH = 0.1464 M[/tex]
- mass KHP = 0.5120 g
[tex]volume NaOH = 16.7 mL = 0.0167 Lmoles NaOH = 2.5071 molM NaOH = 0.1501 M[/tex]
Hence, the molarity of the following is 0.1501M.
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