WebbPoiché le concentrazioni di ioni idronio e idrossido sono pari per acqua pura, avrai. [H3O+] = √10−14 = 10−7M. The pH di acqua pura sarà così. pH = − log([H3O+]) pH = − log(10−7) = 7. Ora supponiamo che tu stia lavorando con a 1.0-L soluzione di acqua pura e ne aggiungi alcuni 10−8M soluzione di acido cloridrico. Webb14 mars 2024 · The pH of 10⁻⁸ HCl solution taking into account the h+ produced by water will be 6.98. Explanation: 10⁻⁸ HCl will dissociates to give 10⁻⁸ H⁺ ions Similarly water will dissociates to give equal concentration of H⁺ and OH⁻ ions H₂O = H⁺ + OH⁻ [H⁺] = [OH⁻] = x lets say Hence in the solution, net concentration of H⁺ ions will be (x + 10⁻⁸)

The pH of \[{10^{ - 8}}\] M HCl solution is: - Vedantu

Webb10 apr. 2024 · pH = 6.99 Hence, the pH value of 1 × 10-8 (M) HCl is 6.98. Download Solution PDF Latest CG TET Updates Last updated on Jan 31, 2024 The CG TET Revised Results were released on January 17, 2024! The CG TET Answer Key (Provisional) for the exam held on September 18th, 2024 has been released. im look for lawn mower

At 25∘ C the pH of a 10 8 molar solution of HCl in water isA. Between 6

WebbWhat is the pH of a 10 - 10 M solution of HCl? Tutorial pH of Solution with Very Dilute Acid The negative log of 10 - 10 gives a pH of 10 -- which means that the solution is basic. Does it make sense that a very dilute acid would have the pH of a base? WebbEvaluate the pH of 7.8 M HCL Solution easily using the pH Calculator and get the resultant pH Value instantaneously. Ex: 0.03, 0.45, 0.34 or 0.0034 Concentration Step by Step Solution finding the pH of 7.8 M HCL Solution pH is defined as the negative of the base-ten logarithm of the molar concentration of hydrogen ions present in the solution. WebbThe pH of 10−8M H Cl solution is 5687 61 IIT JEE IIT JEE 1981 Equilibrium Report Error A 8. B more than 8 C between 6 and 7 D slightly more than 7 Solution: From H 2O,[H 2] = 1×10−7M From H C I,[H +] = 1×10−8M Total [H +] = (1×10−7 +1×10−8)M = (1× 10−7 + 0.1×10−7)M = 1.1× 10−7M = 1.1× 10−7M pH = −log(1.1×10−7) = 6.9586 i’m looking forward to hearing from you soon