Strong Acids and Bases, pH, pOH - AP Chemistry

1. State the formula for pH in terms of hydronium concentration.

   $\mathrm{pH}=-\log[\mathrm{H_3O^+}]$. Negative log converts small $[\mathrm{H_3O^+}]$ values to manageable pH scale.

2. What is the $\text{pH}$ of a $0.010\,\text{M}$ strong monoprotic acid?

   $\text{pH}=2.00$. Monoprotic: n=1, so [H⁺]=0.010 M; pH=-log(0.010)=2.00.

3. What is the $[\text{OH}^-]$ in a solution with $[\text{H}^+]=1.0\times10^{-4}\,\text{M}$ at $25^\circ\text{C}$?

   $[\text{OH}^-]=1.0\times10^{-10}\,\text{M}$. Use Kw=[H⁺][OH⁻]=1.0×10⁻¹⁴; [OH⁻]=1.0×10⁻¹⁴/1.0×10⁻⁴.

4. What is $[\mathrm{H_3O^+}]$ for a $0.030\ \mathrm{M}$ solution of $\mathrm{HCl}$?

   $[\mathrm{H_3O^+}]=0.030\ \mathrm{M}$. HCl is monoprotic strong acid, so $[\mathrm{H_3O^+}]$ equals molarity.

5. What is the $[\text{H}^+]$ when $\text{pH}=5.00$?

   $[\text{H}^+]=1.0\times10^{-5}\,\text{M}$. Apply [H⁺]=10⁻ᵖᴴ=10⁻⁵=1.0×10⁻⁵ M.

6. What is the formula for converting $\text{pH}$ to $[\text{H}^+]$?

   $[\text{H}^+]=10^{-\text{pH}}$. Inverse log operation converts pH back to concentration.

7. What is the formula that relates $\text{pOH}$ to $[\text{OH}^-]$?

   $\text{pOH}=-\log[\text{OH}^-]$. Negative log converts hydroxide ion concentration to pOH scale.

8. What assumption is used for a strong base to relate $[\text{OH}^-]$ to molarity?

   $[\text{OH}^-]\approx n\,C_{\text{base}}$. Strong bases fully dissociate; n is OH⁻ ions per base molecule.

9. What is the pOH of a solution with $[\mathrm{H_3O^+}]=1.0\times10^{-9}\ \mathrm{M}$ at $25^\circ\mathrm{C}$?

   $\mathrm{pOH}=5.00$. $\mathrm{pH}=-\log(1.0\times10^{-9})=9$; $\mathrm{pOH}=14-9=5$.

10. What does $n$ represent in $[\text{H}^+]\approx n\,C_{\text{acid}}$ for a strong acid?

    $n=\text{moles of }\text{H}^+\text{ produced per mole of acid}$. Counts H⁺ ions released per acid molecule (e.g., 2 for H₂SO₄).

11. What is the $\text{pH}$ of a $0.10\,\text{M}$ strong base that provides $3\,\text{OH}^-$ per formula unit?

    $\text{pH}=13.48$. n=3, so [OH⁻]=3×0.10=0.30 M; pOH=0.52; pH=14-0.52=13.48.

12. State the formula for pOH in terms of hydroxide concentration.

    $\mathrm{pOH}=-\log[\mathrm{OH^-}]$. Negative log converts small $[\mathrm{OH^-}]$ values to manageable pOH scale.

13. What assumption is used for a strong acid to relate $[\text{H}^+]$ to molarity?

    $[\text{H}^+]\approx n\,C_{\text{acid}}$. Strong acids fully dissociate; n is H⁺ ions per acid molecule.

14. What is the relationship between pH and pOH at $25^\circ\mathrm{C}$?

    $\mathrm{pH}+\mathrm{pOH}=14.00$. At $25°\mathrm{C}$, this sum equals $-\log(K_w)=14.00$.

15. What is the pOH of a $0.015\ \mathrm{M}$ solution of $\mathrm{Ba(OH)_2}$?

    $\mathrm{pOH}=1.52$. $\mathrm{pOH}=-\log(0.030)=-\log(3\times10^{-2})=1.52$.

16. What is the pH of a $1.0\times10^{-3}\ \mathrm{M}$ solution of $\mathrm{HNO_3}$?

    $\mathrm{pH}=3.00$. $\mathrm{pH}=-\log(1.0\times10^{-3})=3.00$ for strong acid HNO₃.

17. What is the pH of a $2.0\times10^{-5}\ \mathrm{M}$ solution of $\mathrm{HBr}$?

    $\mathrm{pH}=4.70$. $\mathrm{pH}=-\log(2.0\times10^{-5})=5-\log(2)=4.70$.

18. What is the $\text{pOH}$ when $\text{pH}=3.25$ at $25^\circ\text{C}$?

    $\text{pOH}=10.75$. Use pH+pOH=14.00; pOH=14.00-3.25=10.75.

19. Identify the pH of pure water at $25^\circ\mathrm{C}$.

    $\mathrm{pH}=7.00$. Neutral water has $[\mathrm{H_3O^+}]=[\mathrm{OH^-}]=1.0\times10^{-7}\ \mathrm{M}$.

20. What is $[\mathrm{OH^-}]$ for a $0.015\ \mathrm{M}$ solution of $\mathrm{Ba(OH)_2}$?

    $[\mathrm{OH^-}]=0.030\ \mathrm{M}$. Ba(OH)₂ releases 2 OH⁻ per unit: $0.015\times^2=0.030\ \mathrm{M}$.

21. What is the $[\text{OH}^-]$ when $\text{pOH}=9.00$?

    $[\text{OH}^-]=1.0\times10^{-9}\,\text{M}$. Apply [OH⁻]=10⁻ᵖᴼᴴ=10⁻⁹=1.0×10⁻⁹ M.

22. What is the pH of a $4.0\times10^{-2}\ \mathrm{M}$ solution of $\mathrm{NaOH}$ at $25^\circ\mathrm{C}$?

    $\mathrm{pH}=12.60$. $\mathrm{pH}=14.00-\mathrm{pOH}=14.00-1.40=12.60$.

23. What is the formula that relates $\text{pH}$ to $[\text{H}^+]$?

    $\text{pH}=-\log[\text{H}^+]$. Negative log converts hydrogen ion concentration to pH scale.

24. Identify the assumption used for strong acids when finding $[\mathrm{H_3O^+}]$.

    Strong acids dissociate $\approx100\%$: $[\mathrm{H_3O^+}]\approx$ acid molarity. Complete dissociation means initial acid concentration equals $[\mathrm{H_3O^+}]$.

25. Identify the assumption used for strong bases when finding $[\mathrm{OH^-}]$.

    Strong bases dissociate $\approx100\%$: $[\mathrm{OH^-}]\approx$ base molarity $\times$ stoichiometry. Complete dissociation; multiply by OH⁻ per formula unit.

26. What is $K_w$ at $25^\circ\text{C}$ for water?

    $K_w=[\text{H}^+][\text{OH}^-]=1.0\times10^{-14}$. Water's ion product constant at 25°C.

27. What is the formula for converting $\text{pOH}$ to $[\text{OH}^-]$?

    $[\text{OH}^-]=10^{-\text{pOH}}$. Inverse log operation converts pOH back to concentration.

28. What is the pH of a $3.0\times10^{-4}\ \mathrm{M}$ solution of $\mathrm{Ca(OH)_2}$ at $25^\circ\mathrm{C}$?

    $\mathrm{pH}=10.78$. $[\mathrm{OH^-}]=2\times^3.0\times10^{-4}=6.0\times10^{-4}$; $\mathrm{pOH}=3.22$; $\mathrm{pH}=10.78$.

29. What is the $\text{pOH}$ of a $0.0010\,\text{M}$ strong monohydroxide base?

    $\text{pOH}=3.00$. Monohydroxide: n=1, so [OH⁻]=0.0010 M; pOH=-log(0.0010)=3.00.

30. What is $[\mathrm{H_3O^+}]$ for a solution with $\mathrm{pH}=2.50$?

    $[\mathrm{H_3O^+}]=10^{-2.50}=3.2\times10^{-3}\ \mathrm{M}$. Use $[\mathrm{H_3O^+}]=10^{-\mathrm{pH}}=10^{-2.50}$.

31. What is the pH of a $0.015\ \mathrm{M}$ solution of $\mathrm{Ba(OH)_2}$ at $25^\circ\mathrm{C}$?

    $\mathrm{pH}=12.48$. $\mathrm{pH}=14.00-\mathrm{pOH}=14.00-1.52=12.48$.

32. What is the $\text{pH}$ of a $0.0050\,\text{M}$ strong base that provides $2\,\text{OH}^-$ per formula unit?

    $\text{pH}=12.00$. n=2, so [OH⁻]=2×0.0050=0.010 M; pOH=2.00; pH=14-2=12.00.

33. What does $n$ represent in $[\text{OH}^-]\approx n\,C_{\text{base}}$ for a strong base?

    $n=\text{moles of }\text{OH}^-\text{ produced per mole of base}$. Counts OH⁻ ions released per base molecule (e.g., 2 for Ba(OH)₂).

34. What is the $\text{pH}$ when $\text{pOH}=1.60$ at $25^\circ\text{C}$?

    $\text{pH}=12.40$. Use pH+pOH=14.00; pH=14.00-1.60=12.40.

35. What is the pOH of a $4.0\times10^{-2}\ \mathrm{M}$ solution of $\mathrm{NaOH}$?

    $\mathrm{pOH}=1.40$. $\mathrm{pOH}=-\log(4.0\times10^{-2})=2-\log(4)=1.40$.

36. What is the $\text{pH}$ of a $0.020\,\text{M}$ strong diprotic acid?

    $\text{pH}=1.40$. Diprotic: n=2, so [H⁺]=2×0.020=0.040 M; pH=-log(0.040)=1.40.

37. State the expression for $K_w$ and its value at $25^\circ\mathrm{C}$.

    $K_w=[\mathrm{H_3O^+}][\mathrm{OH^-}]=1.0\times10^{-14}$. Water autoionization constant at $25°\mathrm{C}$.

38. What is $[\mathrm{OH^-}]$ for a solution with $\mathrm{pOH}=5.20$?

    $[\mathrm{OH^-}]=10^{-5.20}=6.3\times10^{-6}\ \mathrm{M}$. Use $[\mathrm{OH^-}]=10^{-\mathrm{pOH}}=10^{-5.20}$.

39. What is the pH of a solution with $[\mathrm{OH^-}]=1.0\times10^{-12}\ \mathrm{M}$ at $25^\circ\mathrm{C}$?

    $\mathrm{pH}=2.00$. $\mathrm{pOH}=-\log(1.0\times10^{-12})=12$; $\mathrm{pH}=14-12=2$.

40. What is the relationship between $\text{pH}$ and $\text{pOH}$ at $25^\circ\text{C}$?

    $\text{pH}+\text{pOH}=14.00$. At 25°C, the sum always equals 14 due to water's ion product.

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