1.

Calculate the [OH-] of a solution whose pH is 9.62.

Answer»

pH + pOH = 14 pOH = 14 – pH = 14 – 9.62 = 4.38 

pOH = 4.38 = – log10 [OH ]; 4.38 = – log10 [OH ]

log10[ OH ] = – 4.38 ; [OH] = antilog (–4.38) 

= antilog (5 – 4.38 – 5) = antilog (0.62 – 5) 

= antilog (0.62) x 10–5 = 4.169 × 10–5 mol dm–3.



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