Respuesta :

log1.429V=3.20  volume (to the nearest 0. 1 ml) of 4. 70-m NaOH must be added to 0. 700 l of 0. 250-m HNO2 to prepare a ph = 3. 20 buffer

Solution ;

pH= pKa +log {[ NaOH ] / [ HNO2 ]​}

3.20=3.80+log{(V×4.70)/ ( 0.700×0.250)}

3.20=log1.429V

Volume will be 3.20=log1.429V

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