Home
Class 11
CHEMISTRY
A student spills a small amount of "NaOH...

A student spills a small amount of "NaOH" into a vessel containing "200" L of water, which increases the pH from 7 to 9. How many grams of `H_(2)SO_(4)` must be added further to bring the pH back to 7 ?

Promotional Banner

Similar Questions

Explore conceptually related problems

100mL of a buffer solution contains 0.1M each of weak acid HA and salt NaA . How many gram of NaOH should be added to the buffer so that it pH will be 6 ? (K_(a) of HA = 10^(-5)) .

100 mL of a buffer solution contains 0.1 M each of weak acid HA and salt NaA . How many gram of NaOH should be added to the buffer so that it pH will be 6 ? ( K_(a) of HA=10^(-5) ).

An environment chemist needs a carbonate buffer of pH=10 to study the effects of the acidification of limestones rich solis.How many grams of Na_(2)CO_(3) must be added to 1.5L of freshly prepared 0.2M NaHCO_(3) to make the buffer.? For H_(2)CO_(3),K_(a_(1))=4.7xx10^(-7),K_(a_(2))=4.7xx10^(-11)

In 729 litres solution of acid and water, the ratio of acid and water 7:2. How many litres of water must be added to it to get the solution in which the ratio of acid and water is 5:3 ?

a. What amount of H_(2)SO_(4) must be dissolved in 500mL of solution to have a pH of 2.15 ? b. What amount of KOH must be dissolved in 200mL of solution to have a pH of 12.3 ? c. What amount of ca(OH)_(2) must be dissolved in 100mL of solution to have a pH of 13.85 ?

Using molarity as a conversion factor: An experiment calls for the addition to a reaction vessel of 0.200 g of sodium hydroxide (NaOH) in aqueous solution. How many milliliters of 0.200 M NaOH should be added? Strategy: According to equating 2.13 M=n_(solute)/V_(L) First we need to convert grams NaOH to moles NaOH , because molarity relates moles of solute to volume of solution. Then, we convert moles NaOH to litres of solution, using the molarity as a conversion factor. Here, 0.200 M means that 1 L of solution contains 0.200 moles of solute (NaOH) , so the conversion factor is: (1L soln)/(0.200 mol NaOH) (Converts mol NaOH to L soln)