The volume of H₃PO₄ : 13.33 ml
<h3>Further explanation</h3>
Given
0.003 M Phosphoric acid-H₃PO₄
40 ml of 0.00150 M Calcium hydroxide-Ca(OH)₂
Required
Volume of H₃PO₄
Solution
Acid-base titration formula
Ma. Va. na = Mb. Vb. nb
Ma, Mb = acid base concentration
Va, Vb = acid base volume
na, nb = acid base valence (amount of H⁺/OH⁻)
H₃PO₄⇒3H⁺ + PO₄³⁻ ⇒ 3 H⁺ = valence = 3
Ca(OH)₂⇒Ca²⁺ + 2OH⁻⇒ 2 OH⁻ = valence = 2
Input the value :
a = H₃PO₄, b = Ca(OH)₂
0.003 x Va x 3 = 0.0015 x 40 x 2
Va = 13.33 ml
Answer:
The answer to your question is 245 grams
Explanation:
Data
Volume 6.5 L
Molarity = 0.34
mass of CaCl₂ = ?
Process
1.- Calculate the molar mass of CaCl₂
molar mass = (1 x 40) + (2 x 35.5)
= 40 + 71
= 111 g
2.- Convert the grams to moles
111 g of CaCl₂ -------------- 1 mol
x ---------------0.34 mol
x = (0.34 x 111) / 1
x = 37.74 g
3.- Calculate the total mass
37.74 g ------------------ 1 L
x ------------------ 6.5 L
x = (6.5 x 37.74) / 1
x = 245.31
In 5.70 mol of Hafnium there are 34,326
