The dissociation of water at $25 \, ^\circ C$ is $1.9 \times 10^{-7} \%$ and the density of water is $1.0 \, g/cm^3$. The ionisation constant of water is

At $25\, ^oC$,the dissociation constant of $CH_3COOH$ and $NH_4OH$ in aqueous solution are almost the same. The $pH$ of a $0.01\, N$ $CH_3COOH$ solution is $4.0$ at $25\, ^oC$. The $pH$ of a $0.01\, N$ $NH_4OH$ solution at the same temperature would be:

Calculate the $pH$ of the resultant mixtures:
$(a)$ $10 \, mL$ of $0.2 \, M \, Ca(OH)_{2} + 25 \, mL$ of $0.1 \, M \, HCl$
$(b)$ $10 \, mL$ of $0.01 \, M \, H_{2}SO_{4} + 10 \, mL$ of $0.01 \, M \, Ca(OH)_{2}$
$(c)$ $10 \, mL$ of $0.1 \, M \, H_{2}SO_{4} + 10 \, mL$ of $0.1 \, M \, KOH$

The pairs of compounds which cannot exist together in aqueous solution are :
$I. NaH_2PO_4$ and $Na_2CO_3$ $II. Na_2CO_3$ and $NaHCO_3$
$III. NaOH$ and $NaH_2PO_4$ $IV. NaHCO_3$ and $NaOH$

Assign $A, B, C, D$ from the given type of reaction.
$Cu(OH)_2 \downarrow + 4NH_3(soln.) \longrightarrow [Cu(NH_3)_4]^{2+} + 2OH^{-}$

The hydrogen ion concentration in a mixture of $10 \ mL$ of $0.1 \ M$ $H_2SO_4$ and $10 \ mL$ of $0.1 \ M$ $KOH$ solution in water,is . . . . . . $M$.