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calculate [h ] and ph for the following solutions a) 0.010 m hno3 b) 0.035 m koh c) 0.030 m hcl d) 3.0 m hcl e) 0.010 m [(ch3)4n ]oh-
calculate [h ] and ph for the following solutions a) 0.010 m hno3 b) 0.035 m koh c) 0.030 m hcl d) 3.0 m hcl e) 0.010 m [(ch3)4n ]oh-
a) ThepHof 0.010 m HNO₃ is 2.b) ThepHof 0.035m KOH is 12.54.c) ThepHof 0.030 mhclis 1.52d) ThepHof 3.0 m hcl is 0.52.e) ThepHof 0.010 m [(CH₃)₄N]OH⁻ is 12.0.To calculate the [H⁺] and pH for the given solutions, we can use the equations:[H⁺] = 10^(-pH)pH = -log[H⁺]a) 0.010 M HNO₃[H⁺] = 0.010 MpH = -log(0.010) = 2b) 0.035 M KOHKOHdissociatesinto K+ and OH⁻, so the [OH⁻] is equal to the concentration of KOH.[OH⁻] = 0.035 MTo find the [H⁺], we can use theequation:[H⁺][OH⁻ = 1.0 x 10^(-14)[H⁺] = 1.0 x 10^(-14) / 0.035[H⁺] = 2.86 x 10^(-13) MpH = -log(2.86 x 10^(-13)) = 12.54c) 0.030 M HCl[H⁺] = 0.030 MpH = -log(0.030) = 1.52d) 3.0 M HCl[H⁺] = 3.0 MpH= -log(3.0) = 0.52e) 0.010 M [(CH₃)₄N]OH⁻[(CH₃)₄N]OH⁻ dissociates into (CH₃)₄N+ and OH, so the [OH⁻] is equal to theconcentrationof [(CH₃)₄N]OH⁻.[OH⁻] = 0.010 MTo find the [H⁺], we can use the equation:[H⁺][OH⁻] = 1.0 x 10^(-14)[H⁺] = 1.0 x 10^(-14) / 0.010[H⁺] = 1.0 x 10^(-12) MpH = -log(1.0 x 10^(-12)) = 12.0To know more aboutpHclick on below link:brainly.com/question/15289741##SPJ11...