Answer is: volume of H₂SO₄ is 42.1 mL.<span>
Chemical reaction: H</span>₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.<span>
c(H</span>₂SO₄) = 0,4567 M = 0,4567 mol/L.<span>
V(NaOH) = 30 mL </span>÷ 1000 mL/L <span>= 0,03 L.
c(NaOH) = 0,321 M = 0,321 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,321 mol/L · 0,030 L.
n(NaOH) = 0,00963 mol.
From chemical reaction: n(H</span>₂SO₄) : n(NaOH) = 1 : 2.<span>
n(H</span>₂SO₄) = 0,01926 mol.<span>
V(H</span>₂SO₄) = n(H₂SO₄) ÷ c(H₂SO₄).<span>
V(H</span>₂SO₄) = 0,01926 mol ÷ 0,4567 mol/L.<span>
V(H</span>₂SO₄<span>) = 0,0421 L = 42,1 mL.</span>
Weak base: [OH⁻] = √Kb.C
pKb = 4.2
c = concentration
MM Amphetamine (C9H13N) = 135.21 g/mol
c = 215 mg/L = (0.215 g : 135,21 g/mol) / L = 0.00159 mol/L = 1.59 x 10⁻³ mol/L
![\tt [OH^-]=\sqrt{10^{-4.2}\times 1.59\times 10^{-3}}=3.17\times 10^{-4}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7B10%5E%7B-4.2%7D%5Ctimes%201.59%5Ctimes%2010%5E%7B-3%7D%7D%3D3.17%5Ctimes%2010%5E%7B-4%7D)
pOH = 4 - log 3.17
pH = 14 - (4 - log 3.17)
pH = 10 + log 3.17 = 10.50
I believe it’s A but I could be wrong
