Which change causes the pH of 2.0 cm3 of 1.0×10−1 mol dm−3 HNO31.0 \times 10^{-1}\text{ mol dm}^{-3}\text{ HNO}_31.0×10−1 mol dm−3 HNO3 to be tripled at 298 K? (Take Kw=1.0×10−14 mol2 dm−6K_\text{w} = 1.0 \times 10^{-14}\text{ mol}^2\text{ dm}^{-6}Kw=1.0×10−14 mol2 dm−6 at 298 K)
adding 18 cm318\text{ cm}^318 cm3 of water
adding 198 cm3198\text{ cm}^3198 cm3 of water
adding 1998 cm31998\text{ cm}^31998 cm3 of water
adding 2.0 cm32.0\text{ cm}^32.0 cm3 of 1.0×10−1 mol dm−3 NaOH1.0 \times 10^{-1}\text{ mol dm}^{-3}\text{ NaOH}1.0×10−1 mol dm−3 NaOH
