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PH of acidic buffer = pKa + log [CH₃COONa - HCl] / [CH₃COOH + HCl]
pKa of CH₃COOH = 4.74
Concentration of acetic acid in buffer = 2.0 M
Concentration of sodium acetate = 1.0 M
Concentration of HCl must add = x
pH = 4.74 + log (1-x) / (2+x) = 4.11
x = concentration of HCl must be added = 0.43 M
number of moles of HCl = M * V = 0.43 * 1 = 0.43 mol
mass of HCl must be added = 0.43 * 36.5 = 15.7 g
Answer:hey did you ever get an answer? I need it for this exam
Explanation:
The joules required to heat 2L of water in a pot from 20 c to the boiling point of water is calculated using the following formula
Q= MC delta T
M = mass = density x volume( 2 x 1000= 2000ml)
M = 1g/ml x2000 ml = 2000g
C = specific heat capacity = 4.18 g/c
delta T = change in temperature = 100 c ( boiling point of water) - 20 c = 80 c
Q is therefore = 2000 g x 4.18 j/g c x 80c = 668800j
Answer:
K ^+ , CO3 ^2-
Explanation:
The compound is potassium trioxocarbonate(IV).
It contains cation (potassium ion) and acid radical ( trioxocarbonate (IV) ion).
Since K is in group 1 of the periodic table, it loses one electron to form ion i.e K^1. trioxocarbonate IV ion has a charge of 2-.and so the ions of the compound are as shown in the answer above.
