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Ionic Equilibrium question

2025 · 4 Apr · Shift 1 · Q21
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Ionic Equilibrium question

2025 · 4 Apr · Shift 1 · Q21

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
The pH of a 0.01 M weak acid HX(Ka=4×10−10)\mathrm{HX}\left(\mathrm{K}_a=4 \times 10^{-10}\right)HX(Ka​=4×10−10) is found to be 5 . Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6 . The new concentration of the diluted weak acid is given as x×10−4Mx \times 10^{-4} \mathrm{M}x×10−4M. The value of xxx is ‾\underline{\hspace{2cm}}​ (nearest integer)
Numerical answer
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