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Another chem questions to be checked
Calculate the pH of the sol. formed when 3.5g of impure NaOH (98.7%) is dissolved in water and made up to 100cm^3,and then 25cm^3 of 0.35M dibasic acid is added?so i did
3.5/23+16+1= 0.0875
0.0875*(1-0.13)= 0.076125mols
then i did
25/1000 * 0.35x2 = 0.0175
left over = 0.058625mols (OH)
then used kw go get H* conc of 2.13x10^-14 and ph of 13.67
can some1 check my calc please
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