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What volume of 0.2 M NaOH in L) solution should be mixed to 2 of 0.5M NaOH solution so that 300ml of final solution is completely neutralised by 20 ml of 2 M H3PO4 solution. [Assuming 100%dissociation]
11 months ago

20ml of 2M H3PO4 ⇒ mol H3PO4 = 2 × 0.02 = 0.04 mol ⇒ ΣnH+ = 3 × moles of H3PO4 = 0.04 × 3 = 0.12 mol
we have:
H⁺ + OH⁻ –-–-→ H2O
0.12–-→ 0.12 mol
therefor In 300 ml NaOH (after mixed) , will have 0.12 mol OH⁻, because NaOH → Na⁺ + OH⁻
⇒ moles of NaOH is 0.12 mol (in 300 ml NaOH after mixed ) (ℂ )

let x is vol. of 0.2M NaOH ( in liters), ⇒ moles NaOH in x (litters) 0.2M is 0.2x mol
500 ml = 0.5ℓ of 0.5 M NaOH sol ⇒ moles NaOH in it is 0.5 × 0.5 = 0.25 mol
⇒ In (0.5 + x)ℓ NaOH sol, Σn(NaOH) = 0.25 + 0.2x (mol)
_________________________________(0.25 + 0.2x ) × 0.3
⇒ in 300 ml = 0.3ℓ, moles of NaOH =▬▬▬▬▬▬▬▬▬▬▬ mol
__________________________________0.5 + x
___________(0.25 + 0.2x ) × 0.3
From (ℂ ) ⇒ ▬▬▬▬▬▬▬▬▬ = 0.12 ⇒ x = 0.25 ℓ = 250 ml
______________0.5 + x

the answer is 250
11 months ago
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