This is an acid base reaction and the chemical equation for the above reaction is as follows;
KOH + HClO₄ ---> KClO₄ + H₂O
the stoichiometry of acid to base is 1:1
KOH is a strong base and HClO₄ is a strong acid therefore they both ionize completely into their respective ions
Number of KOH moles - 0.723 M/1000 mL/L x 25.0 mL = 0.018 mol
Number of HClO₄ moles - 0.273 M/1000 mL/L x 50 mL = 0.013 mol
since acid and base react completely, 0.013 mol of acid reacts with 0.013 mol of base.
The excess base remaining is - 0.018 - 0.013 = 0.005 mol
total volume of solution = 25.0 mL + 50.0 mL = 75.0 mL
[OH⁻] = 0.005 mol/0.075 L = 0.067 M
pOH = -log[OH⁻]
pOH = -log(0.067 M)
pOH = 1.17
pOH + pH = 14
Therefore pH = 14 - 1.17 = 12.83
by knowing pH we can calculate the [H₃O⁺]
pH = -log [H₃O⁺]
[H₃O⁺] = antilog[-12.83]
[H₃O⁺]= 1.47 x 10⁻¹³ M
1)
HI(aq) → H⁺(aq) + I⁻(aq)
So this is an Arrhenius acid because it releases H⁺.
2)
LiOH(s) → Li⁺ + OH⁻
So this is an Arrhenius base because it releases OH⁻
Answer:
750mmHg
Explanation:
The following data were obtained from the question:
T1 = 127°C = 127 +273 = 400K
T2 = 27°C = 27 +273 = 300K
P1 = 1000mmHg
P2 =?
P1/T1 = P2/T2
1000/400 = P2 /300
Cross multiply
400 x P2 = 1000 x 300
Divide both side by 400
P2 = (1000 x 300)/400
P2 = 750mmHg
Therefore, the new pressure after cooling is 750mmHg