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 number of milliliters of a 1M is 402mL
Explanation:
The computation of the number of milliliters could be determined by using the following formula
As we know that
where,
V_1 and V_2 are the starting and final volumes
And, the M_1 and M_2 are the starting and the final molarities
Now the V_1 is
So, the V_1 is 402mL
Hence, the number of milliliters of a 1M is 402mL
Answer: 0.0738 M
Explanation:
1) Chemical equation:
<span>HNO₃ + LiOH → LiNO₃ + H₂O
</span>
2) Mole ratios:
1 mol HNO₃ neutralizes 1 mol LiOH
3) Number of moles of H⁺ from HNO₃:
M = n / V ⇒ n = M × V = 0.050 liter × M
4) Number of moles of OH⁻ from LiOH
M = n / V ⇒ n = M × V = 0.03690 liter × 0,100 M
5) Equivalence point:
number of moles of H⁺ = number of moles of OH⁻
⇒ 0.0500 × M = 0.03690 × 0.100
⇒ M = 0.03690 × 0.100 / 0.0500 = 0.0738 M
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B 404 ml hahahahhahahahahahahhahahahahahahhaahhahahahahaha
