Bibliii KOH as God notes is the strong base potassium hydroxide, potassium is denoted by the symbol K since it’s called kalium in Latin. To approach this question, I’d take -log10( ) of the concentration of KOH they gave you to get the pOH, and then rearrange the formula pOH + pH = pKw = 14 for pH
For 17, hydrocyanic acid is a very weak acid judging from its low Ka magnitude so set up an ICE table as follows
Initial [HCN] = 0.227 M, initial [CN-] = 0 M, initial [H+] = 0 M
Change [HCN] = -x M, change [CN-] = +x M, change [H+] = +x M
Equilibrium [HCN] = 0.227 - x M, equilibrium [CN-] = x M, equilibrium [H+] = x M
Then equate the magnitude of Ka with your Ka expression
6.2 x 10^-10 = (x M)2 / (0.227 M)*
*In uni chem here in Victoria we’re always allowed to assume that the concentration change of an weak acid is 0, if QCE doesn’t allow it, you’ll have to use the quadratic formula.
Now solving for x should give you [H+], which you should then convert into pH
Ka isn’t the same as pKa, pKa is -log10(Ka) (whenever you see p in chem notation, it means -log10( ), such that pH = -log10(H+) and pOH = -log10(OH-))
18 is a titration question that involves a dilution, always start with a balanced equation and remember that your standard was diluted by a factor of 10.
So you know the original concentration of your standard was 31.8 g / 106 g/mol = 0.3 mol in 500 mL = 0.6 M. Diluting that by a factor 10 (use c1v1 = c2v2, where c1 = 0.6 M, v1 = 5 mL, c2 is your unknown and v2 = 50 mL). Your DILUTED concentration would be 0.06 M
Now you know 50 mL of a 0.06 M Na2CO3 solution reacted with 16.8 mL of H2SO4 solution. The balanced equation is pretty simple since everything is 1:1:1:1:1 (Na2CO3 + H2SO4 —> Na2SO4 + H2O + CO2). Find n(Na2CO3), which will be equivalent to n(H2SO4) according to the balanced molar ratio. Then use c = n/v to find [H2SO4], remember that v must be in L so 16.8 mL = 0.0168 L.
