A52.0-ml volume of 0.35 m ch3cooh (ka=1.8×10−5) is titrated with 0.40 m naoh. calculate the ph after the addition of 17.0 ml of naoh. express your answer numerically.

2.76.

Explanation:

C of acid = [(MV)CH₃COOH - (MV)NaOH] / Vtotal.

C of acid = [(0.35 M)(52.0 mL) - (0.4 M)(17.0 mL)] / (69.0 mL) = 0.165 M.

∵ [H⁺] = √(Ka.C)

∴ [H⁺] = √(1.8 x 10⁻⁵)(0.165 M) = 1.72 x 10⁻³.

∵ pH = - log[H⁺] = - log (1.72 x 10⁻³) = 2.76.

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

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