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Part A Calculate the change in pH that results from adding 0.190 mol NaNO2 to 1.00 L...

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Chemistry

Part A

Calculate the change in pH that results from adding 0.190 molNaNO2 to 1.00 L of 0.190 M HNO2(aq). (Ka(HNO2)=7.2⋅10−4) Expressyour answer using two decimal places.

Part B

Calculate the change in pH that results from adding 0.190 NaNO3to 1.00 L of 0.190 HNO3(aq). Express your answer using two decimalplaces.

Answer & Explanation Solved by verified expert
3.8 Ratings (385 Votes)
Answer Part A we are given 10 L of 0190 M HNO2 Ka 72104 pH before adding the NaNO2 we know HNO2 is weak acid HNO2 H2O H3O NO2 I 0190 0 0 C x x x E 0190x x x Ka H3O NO2 HNO2 72104 xx 0190x 72104 0190x    See Answer
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