Answer:
a) 0.714g of bicarbonate of soda are required.
b) 0.221g of Al(OH)₃ are required
Explanation:
The reactions of HCl with bicarbonate of soda and aluminium hydroxide are:
HCl + NaHCO₃ → H₂O + NaCl + CO₂
3 HCl + Al(OH)₃ → 3H₂O + AlCl₃
The moles of HCl that we need neutralize are:
50mL = 0.050L * (0.17mol / L) = 0.0085 moles HCl
To solve these problem we need to find the moles of the antacid using the chemical reaction and its mass using its molar mass;
<em>a) </em><em>Moles NaHCO₃ = Moles HCl = 0.0085 moles </em>
The mass is -Molar mass NaHCO₃: -84g/mol-
0.0085 moles * (84g / mol) = 0.714g of bicarbonate of soda are required
b) 0.0085 moles HCl * (1mol Al(OH)₃ / 3mol HCl) = 2.83x10⁻³ moles Al(OH)₃
The mass is -Molar mass: 78g/mol-:
2.83x10⁻³ moles Al(OH)₃ * (78g/mol) =
<h3>0.221g of Al(OH)₃ are required</h3>
Answer:
(3) 5.36
Explanation:
Since this is a titration of a weak acid before reaching equivalence point, we will have effectively a buffer solution. Then we can use the Henderson-Hasselbalch equation to answer this question.
The reaction is:
HAc + NaOH ⇒ NaAc + H₂O
V NaOH = 40 mL x 1 L/1000 mL = 0.040 L
mol NaOH reacted with HAc = 0.040 L x 0.05 mol/L = 0.002 mol
mol HAC originally present = 0.050 L x 0.05 mol/L = 0.0025 mol
mol HAc left after reaction = 0.0025 - 0.002 = 0.0005
Now that we have calculated the quantities of the weak acid and its conjugate base in the buffer, we just plug the values into the equation
pH = pKa + log ((Ac⁻)/(HAc))
(Notice we do not have to calculate the molarities of Ac⁻ and HAc because the volumes cancel in the quotient)
pH = -log (1.75 x 10⁻⁵) + log (0.002/0.0005) = 5.36
THe answer is 5.36
Answer:
Atoms form chemical bonds to make their outer electron shells more stable. ... An ionic bond, where one atom essentially donates an electron to another, forms when one atom becomes stable by losing its outer electrons and the other atoms become stable (usually by filling its valence shell) by gaining the electrons.
Explanation:
