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Question: How to find the molarity of acid and base mixtures and tell me the formula also tell me why the minu...

How to find the molarity of acid and base mixtures and tell me the formula also tell me why the minus sin came here

Answer

The detailed explanation above provides the formulas and explanation for the molarity of acid and base mixtures and the reason for the minus sign. This is a descriptive question, not a multiple-choice one.

Explanation

Solution

To find the molarity of acid and base mixtures, we first determine which reactant is in excess after the neutralization reaction. The molarity of the excess species is then calculated by dividing the moles of the excess reactant by the total volume of the mixture.

Formulas for Molarity of Acid and Base Mixtures:

Let:

  • MAM_A, VAV_A, nAn_A be the molarity, volume (in Liters), and number of acidic protons per molecule (basicity) for the acid, respectively.
  • MBM_B, VBV_B, nBn_B be the molarity, volume (in Liters), and number of hydroxide ions per molecule (acidity) for the base, respectively.

  1. Calculate the moles of H+H^+ from the acid:
    Moles of H+H^+ = MA×VA×nAM_A \times V_A \times n_A

  2. Calculate the moles of OH−OH^- from the base:
    Moles of OH−OH^- = MB×VB×nBM_B \times V_B \times n_B

  3. Calculate the total volume of the mixture:
    Vtotal=VA+VBV_{total} = V_A + V_B

  4. Determine the excess species and its molarity:

    • If Moles of H+H^+ > Moles of OH−OH^- (Acid is in excess):
      Moles of excess H+H^+ = (MAVAnA)−(MBVBnB)(M_A V_A n_A) - (M_B V_B n_B)
      Molarity of excess H+H^+ = [H+]excess=(MAVAnA)−(MBVBnB)VA+VB[H^+]_{excess} = \frac{(M_A V_A n_A) - (M_B V_B n_B)}{V_A + V_B}

    • If Moles of OH−OH^- > Moles of H+H^+ (Base is in excess):
      Moles of excess OH−OH^- = (MBVBnB)−(MAVAnA)(M_B V_B n_B) - (M_A V_A n_A)
      Molarity of excess OH−OH^- = [OH−]excess=(MBVBnB)−(MAVAnA)VA+VB[OH^-]_{excess} = \frac{(M_B V_B n_B) - (M_A V_A n_A)}{V_A + V_B}

    • If Moles of H+H^+ = Moles of OH−OH^- (Neutral solution):
      The solution is neutral (pH ≈\approx 7 for strong acid-strong base).

Why the Minus Sign?

The "minus sign" can refer to two different contexts in acid-base chemistry:

  1. In the Molarity Calculation (Subtraction):
    The minus sign in the formulas for excess molarity (e.g., (MAVAnA)−(MBVBnB)(M_A V_A n_A) - (M_B V_B n_B)) simply represents subtraction. It is used to find the difference between the initial amounts of acid and base, thereby determining the net amount of the reactant that remains unreacted (the excess reactant). You always subtract the smaller quantity from the larger quantity to get a positive result for the moles of excess.

  2. In pH/pOH Formulas (Negative Logarithm):
    The most common "minus sign" in acid-base chemistry is found in the definitions of pH and pOH:
    pH=−log⁡10[H+]pH = -\log_{10}[H^+]
    pOH=−log⁡10[OH−]pOH = -\log_{10}[OH^-]

    This negative sign is used for the following reasons:

    • To Convert Small Numbers to Manageable Positive Integers: The concentrations of H+H^+ and OH−OH^- ions in aqueous solutions are often very small (e.g., 10−710^{-7} M). The logarithm of such a small number is a negative number (e.g., log⁡10(10−7)=−7\log_{10}(10^{-7}) = -7). By taking the negative of this logarithm, the pH and pOH values become positive and easier to work with (e.g., −(−7)=7-(-7) = 7).
    • To Create a Convenient Scale: This convention establishes a practical scale (typically 0-14 for pH) that simplifies the expression and comparison of acidity and basicity across a wide range of concentrations.

The "minus sign" in the pH/pOH formulas is thus a deliberate mathematical operation to convert a vast range of concentrations into a compact, positive, and linear scale.