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) × 10-5 

= 4.169 × 10-5 mol dm-3



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