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Question: The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium n...

The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL of 0.10 M sodium iodate is X × 10⁻⁶ mol dm⁻³. The value of X is

Use Solubility product constant (KspK_{sp}) of barium iodate = 1.58 × 10⁻⁹

Answer

3.95

Explanation

Solution

Here's how to solve the solubility problem:

  1. Calculate the initial concentrations of Ba2+Ba^{2+} and IO3−IO_3^− after mixing:

    • Moles of Ba2+Ba^{2+} from Ba(NO3)2Ba(NO_3)_2: 0.200L⋅0.010M=0.0020 moles0.200 L \cdot 0.010 M = 0.0020 \, moles
    • Moles of IO3−IO_3^− from NaIO3NaIO_3: 0.100L⋅0.10M=0.010 moles0.100 L \cdot 0.10 M = 0.010 \, moles
    • Total volume: 0.200L+0.100L=0.300L0.200 L + 0.100 L = 0.300 L
    • [Ba2+]0=0.0020 mol0.300 L=0.00667 M[Ba^{2+}]_0 = \frac{0.0020 \, mol}{0.300 \, L} = 0.00667 \, M
    • [IO3−]0=0.010 mol0.300 L=0.0333 M[IO_3^−]_0 = \frac{0.010 \, mol}{0.300 \, L} = 0.0333 \, M

  2. Set up the equilibrium expression and ICE table:

    Ba(IO3)2(s)⇌Ba2+(aq)+2IO3−(aq)Ba(IO_3)_2(s) \rightleftharpoons Ba^{2+}(aq) + 2IO_3^−(aq)

    Initial: [Ba2+]=0.00667[Ba^{2+}] = 0.00667, [IO3−]=0.0333[IO_3^−] = 0.0333

    Change: [Ba2+]=+s[Ba^{2+}] = +s, [IO3−]=+2s[IO_3^−] = +2s

    Equilibrium: [Ba2+]=0.00667+s[Ba^{2+}] = 0.00667 + s, [IO3−]=0.0333+2s[IO_3^−] = 0.0333 + 2s

    Ksp=[Ba2+][IO3−]2=(0.00667+s)(0.0333+2s)2=1.58×10−9K_{sp} = [Ba^{2+}][IO_3^−]^2 = (0.00667 + s)(0.0333 + 2s)^2 = 1.58 \times 10^{-9}

  3. Approximate and solve for s:

    Since KspK_{sp} is small and common ions are present, assume 's' is negligible compared to 0.00667 and 0.0333. Because [IO3−][IO_3^-] is squared, it has a stronger common ion effect. So, approximate: [Ba2+]=0.00667+s≈0.00667[Ba^{2+}] = 0.00667 + s \approx 0.00667 [IO3−]=0.0333+2s≈0.0333[IO_3^-] = 0.0333 + 2s \approx 0.0333 Ksp=(0.00667+s)(0.0333)2=1.58×10−9K_{sp} = (0.00667 + s)(0.0333)^2 = 1.58 \times 10^{-9} (0.00667+s)=1.58×10−9(0.0333)2=1.422×10−6(0.00667 + s) = \frac{1.58 \times 10^{-9}}{(0.0333)^2} = 1.422 \times 10^{-6} s=1.422×10−6−0.00667=−0.005248s = 1.422 \times 10^{-6} - 0.00667 = -0.005248

    Since s is negative, precipitation occurs.

  4. Determine if precipitation occurs: Qsp=[Ba2+]0[IO3−]02=(0.00667)(0.0333)2=7.41×10−6Q_{sp} = [Ba^{2+}]_0 [IO_3^-]_0^2 = (0.00667)(0.0333)^2 = 7.41 \times 10^{-6} Qsp>KspQ_{sp} > K_{sp}, so precipitation occurs.

  5. Let 'x' be the amount of Ba2+Ba^{2+} that precipitates.

    Ba2+(aq)+2IO3−(aq)→Ba(IO3)2(s)Ba^{2+}(aq) + 2IO_3^−(aq) \rightarrow Ba(IO_3)_2(s) Initial: [Ba2+]0=1/150[Ba^{2+}]_0 = 1/150, [IO3−]0=1/30[IO_3^−]_0 = 1/30 Change: [Ba2+]=−x[Ba^{2+}] = -x, [IO3−]=−2x[IO_3^−] = -2x Equilibrium: [Ba2+]=1/150−x[Ba^{2+}] = 1/150 - x, [IO3−]=1/30−2x[IO_3^−] = 1/30 - 2x Ksp=(1/150−x)(1/30−2x)2=1.58×10−9K_{sp} = (1/150 - x)(1/30 - 2x)^2 = 1.58 \times 10^{-9} CIO3=0.020+2CBaC_{IO_3} = 0.020 + 2C_{Ba}

    CBaCIO32=1.58×10−9C_{Ba} C_{IO_3}^2 = 1.58 \times 10^{-9}

    CIO3=0.020+2CBaC_{IO_3} = 0.020 + 2C_{Ba}

  6. Solve for equilibrium concentrations:

    CBa(0.020+2CBa)2=1.58×10−9C_{Ba} (0.020 + 2C_{Ba})^2 = 1.58 \times 10^{-9}

    Assume CBaC_{Ba} is very small compared to 0.020:

    CBa(0.020)2=1.58×10−9C_{Ba} (0.020)^2 = 1.58 \times 10^{-9}

    CBa=1.58×10−9(0.020)2=3.95×10−6MC_{Ba} = \frac{1.58 \times 10^{-9}}{(0.020)^2} = 3.95 \times 10^{-6} M

Therefore, the solubility of barium iodate is 3.95×10−6 mol dm−33.95 \times 10^{-6} \, mol \, dm^{-3}.

The value of X is 3.95.