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
A mineral found in a bridge is iron
Answer:
Because of their spaceeeee isss different
<span>Activation energy is _____
</span>
<span>an energy barrier between reactants and products
</span>
Answer:
Explanation:
Initial burette reading = 1.81 mL
final burette reading = 39.7 mL
volume of NaOH used = 39.7 - 1.81 = 37.89 mL .
37.89 mL of .1029 M NaOH is used to neutralise triprotic acid
No of moles contained by 37.89 mL of .1029 M NaOH
= .03789 x .1029 moles
= 3.89 x 10⁻³ moles
Since acid is triprotic , its equivalent weight = molecular weight / 3
No of moles of triprotic acid = 3.89 x 10⁻³ / 3
= 1.30 x 10⁻³ moles .
