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Calculate the pH of the resulting solution if 26.0 mL of 0.260 M HCl(aq) is added to (a) 36.0 mL of 0.260 M NaOH(aq).(b) 16.0 mL of 0.360 M NaOH(aq).
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a. there are 10 ml of 0.26 M NaOH excess from reaction.. \[[OH]^{−}=0.260M×10^{−2}L, pOH=−log[OH]^{−},pH=14−pOH\] b. there are 10 ml of 0.26 M HCL excess from reaction.. \[[H]^{+} = 0.260 M \times 10^{-2} L , pH=-\log[H]^{+}\]
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