If you mix two solutions, 24.0 mL of 1.498 M NaCl and 26.0 mL of 1.995 M NaOH....?

 If you mix two solutions, 24.0 mL of 1.498 M NaCl and 26.0 mL of 1.995 M NaOH, calculate the final sodium (Na+ ), chloride (Cl - ) and hydroxide (OH- ) ion concentrations.

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  • 2 weeks ago

    Mixing ions ....

    NaCl and NaOH ionize completely.

    Na+ .... Cl- .... OH-

    [Na+] = (0.0240L x (1.498 mol Na+ / 1L) + 0.026L x (1.995 mol Na+ / 1L)) / 0.0500L = 1.76M Na+

    [Cl-] = 0.0240L x (1.498 mol Na+ / 1L) / 0.0500L = 0.719M Cl-

    [OH-] = 0.0260L x (1.995 mol Na+ / 1L) / 0.0500L = 1.04M OH-

    The answers can only be expressed to three significant digits.

  • 2 weeks ago

    when these two solutions are mixed there is no reaction so we calculate the molarities separately in 50mL of combined volume 

    moles of Cl- = 1.498 mol /L * 0.0240L = 0.0360 mole 

    in the new volume of of 50 mL molarity = 0.0360 moles  / 0.050L = 0.719 M

    moles of OH- = 1.995 moles / liter * 0.026 L = 0.0519 mole

    in 50.0 mL = 0.0519 moles / 0.050L = 1.037 M 

    Moles of Na+ = (0.0360 mole + 0.0519 mole) / 0.050L = 1.76 M

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