V=1 мл = 10⁻³ л
pH = 13
N(a) = 6.02*10²³ моль⁻¹
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N(OH-) = ?
C[H(+)]*C[OH(-)]=10:(-14) моль/л
рОН=-lg[ OH-]=14-pH = 14 - 13 = 1
lg[OH(-)] = -1
[OH(-)] = 10⁻¹ = 0.1 моль/л
N(OH-) = [OH(-)] * N(a) * V(H2O) = 6.02*10²³ * 0.1 * 10⁻³ = 6.02*10¹⁹ ионов в 1 мл