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$\rho H$ of water is 7.0 at 298 K. If water is heated to 350 K, which of the following should be true?

(a) $\rho OH$ will decrease

(b) $\rho OH$ will increase

(c) $\rho OH$ will remain seven

(d) Concentration of $H^+$ ions will increase but that of $OH^-$ will decrease.

1 Answer

  • Water undergoes self-dissociation, also referred to as self-ionization or autodissociation. $H_2O \rightleftharpoons H^+ + OH^-$
  • The concentration of water is essentially constant at any given temperature, and is incorporated into the equilibrium constant: $K_w = [H^+] [OH^-]$
  • $K_w$ is called the ion product for water or the dissociation constant for water.
  • $K_w$, like all equilibrium constants, is temperature dependent.
  • For any given temperature, $[H^+] = [OH^-]$ for water, so water is neutral at all temperatures.
  • For any given temperature, $pH = pOH$ for water.
  • Water is only $pH$=7 at $25^o C (298K)$
  • Water is only $pOH$=7 at $25^o C (298K)$
  • $ pH + pOH = 14$ is only true at $25^o C$ (298K)
Answer: $pOH$ will decrease.
At 298 K, for pure water $pH = pOH = 7 $. On increasing the temperature, the self ionisation of water increases thereby increasing the concentrations of $H_3O^+$ as well as $OH_-$ ions. Consequently, at high temperature, the $pH$ as well as $pOH$ will be < 7.


answered Nov 27, 2013 by mosymeow_1

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