11.An acid-base equilibrium system is created by dissolving 0.15 mol HNO2 inwater and diluting the resulting solution to a volume of 1.0 L. What is the effectof adding 0.020 mol NO2 (aq) to this solution? How will pH change (calculate pHbefore and after the addition. For HNO2 Ka = 4.5 x 104)? How willconcentrations of HNO2 and NO2 at equilibrium change?

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Asked Nov 25, 2019
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11.An acid-base equilibrium system is created by dissolving 0.15 mol HNO2 in
water and diluting the resulting solution to a volume of 1.0 L. What is the effect
of adding 0.020 mol NO2 (aq) to this solution? How will pH change (calculate pH
before and after the addition. For HNO2 Ka = 4.5 x 104)? How will
concentrations of HNO2 and NO2 at equilibrium change?
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11.An acid-base equilibrium system is created by dissolving 0.15 mol HNO2 in water and diluting the resulting solution to a volume of 1.0 L. What is the effect of adding 0.020 mol NO2 (aq) to this solution? How will pH change (calculate pH before and after the addition. For HNO2 Ka = 4.5 x 104)? How will concentrations of HNO2 and NO2 at equilibrium change?

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pH change before ...

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0.15mol Concentration of HNO - 0.15M 1.0L ICE table: HNO +H,O >NO+H,O Initial(M) 0.15 0 10 Change(M) x +x +x Equili.(M) 0.15-x No,HO к, HNO x2 4.5x10 (0.15-x) x-8.2x103 substituting x HNO,(0.15-x) 0.15-8.2x10)0.142M NO:]-x-8.2x10 From ICE table, H,O]-x8.2x10 pH-log H,O log (8.2x10) -2.09

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