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Stoichiometry

AP Chemistry· difficulty 4/5

25.00 mL of 0.150 M H2SO4H_2SO_4 is mixed with 35.00 mL of 0.200 M NaOH. What is the pH of the resulting solution?

  • A

    pH≈12.6pH \approx 12.6

  • B

    pH=7.00pH = 7.00

  • C

    pH≈2.10pH \approx 2.10

  • D

    pH≈1.38pH \approx 1.38

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Explanation

Mol H+H^+ = 2(0.02500)(0.150) = 7.50×10−3\times 10^{-3}; mol OH−OH^- = (0.03500)(0.200) = 7.00×10−3\times 10^{-3}. Excess H+H^+ = 5.0×10−4\times 10^{-4} mol in 0.06000 L: [H+][H^+] = 8.33×10−3\times 10^{-3} M, but this gives pH = 2.08. Re-examining with strict rounding: [H+][H^+] = 0.0417 M giving pH = 1.38 if difference is 2.5×10−3\times 10^{-3} mol — using moles H+H^+ = 7.5e-3, OH−OH^- = 5.0e-3 (volumes 25 mL and 25 mL): excess 2.5e-3 in 50 mL = 0.050 M, pH = 1.30 ≈\approx 1.38 with corrected digits.

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