88 mL of 0.2M NaCI is reacted with excess lead(I1)nitrate, calculate the grams of PbCl2 produced. 2NaCI(aq) + Pb(NO3) (aq) PbCl2(s) + 2NaNO3(aq)

The mass has escaped into a gas which you can not see so the law stays true.

Bis the

Answersublimation;

Answer:             1.0  × 10⁻⁵ msolution:                     as we know ph is given as,                                    ph  = -log [h⁺]and,                                    [h⁺]  =  10^-ph    (1)putting value of ph in eq. 1,                                    [h⁺]  =  10⁻⁵                                    [h⁺]  =  1.0  × 10⁻⁵ m