Which change causes the pH of 10 cm3 of 1.0×10−2 mol dm−3 HCl1.0 \times 10^{-2}\text{ mol dm}^{-3}\text{ HCl}1.0×10−2 mol dm−3 HCl to be doubled at 298 K? (Take Kw=1.0×10−14K_{\text{w}} = 1.0 \times 10^{-14}Kw=1.0×10−14 at 298 K)
adding 90 cm390\text{ cm}^390 cm3 of water
adding 990 cm3990\text{ cm}^3990 cm3 of water
adding 10 cm310\text{ cm}^310 cm3 of 1.0×10−2 mol dm−3 NaOH1.0 \times 10^{-2}\text{ mol dm}^{-3}\text{ NaOH}1.0×10−2 mol dm−3 NaOH
adding 10 cm310\text{ cm}^310 cm3 of 1.0×10−2 mol dm−3 HCl1.0 \times 10^{-2}\text{ mol dm}^{-3}\text{ HCl}1.0×10−2 mol dm−3 HCl
