NEED HELP ASAP due tommorow been trying to figure this out all week its for science fair compotition

Question

hello I have a project due tommorow and I need help with figuring out how much sulfuric acid I need to put into 2.5oz of water to lower it only 1 pH level and I need to find out the same to higher it with baking soda the things i know about the sulfuric acid are -98.079 g/mol -1.84 g/cm cubed -0.2N -98% baking soda one bag of arm and hammer pure baking soda so with that being said how much sulfuric acid and baking soda to i use to raise and lower the pH by just one?

Answer 1

L water = 2.5 fl oz x 1 gal/128 fl oz  x  3.78541 L/gal = 0.07393 L water

 

x = L H₂SO₄  needed to add to water       

 

pH of 98.079 M H₂SO₄          Test water pH - assume pH 7   and to reduce the pH to 6                     

 

pH = - log[H⁺]            Need pH = 6 = -log[H⁺]    and   10^{6  =} 1/[H⁺]    therefore, [H⁺] = 10^-6 M needed

 

 

 Let x = L of acid added to water

 

 (x*98.079 mol/L acid)/(x L acid + 0.07393 L water)  =  10^-6 mol/L   

 

take out units,

 

98.079*x/(x + 0.07393) = 10^-6        (solve for x)

 

98.079*x = 10^-6*(x + 0.07393) = 10^-6*x + 0.007393       (combine like terms)

 

98.079*x = 0.007393

 

x = 0.007393/98.079 = 7.5378 x 10^-5 L  = 7.5378 x 10^-2 mL acid  

 

or 1 eye drop/(0.05 mL) x 7.5378 x 10^-2 mL acid = 1.5 eye drops 98.079 M acid needed to add to 2.5 fl oz water

 

To raise 2.5 oz of water (at pH 7) to pH 8