# chemistry

posted by .

If 10.0 mL of 0.250 mol/L NaOH(aq) is added to 30.0 mL of 0.17 mol/L HOCN(aq), what is the pH of the resulting solution?

• chemistry -

NaOH + HOCN ==> NaOCN + HOH
mols NaOH to begin = M x L = 0.250 x 0.010 = ??

mols HOCN to begin = M x L = 0.17 x 0.030 = ??

There is more HOCN than NaOH. The difference in mols is how much HOCN is left unreacted. The NaOCN formed is the amount of the lesser chemical (in this case NaOH). So you have a buffer formed consisting of a weak acid (HOCN) and its salt (NaOCN).
Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)]
Post your work if you get stuck.

## Similar Questions

1. ### CHEMISTRY FOR DR. BOB or anyone else

A hypothetical weak acid HA, was combined with NaOH in the following proportions: 0.20 mol HA, 0.08 mol NaOH. The mixture was then diluted to a total volume of 1L, and the pH measured. (a) If pH=4.80, what is the pKa of the acid (b) …
2. ### Chemistry

Use the dilution relationship (Mi x Vi = Mf x Vf) to calculate the volume of 0.500 M NaOH needed to prepare 500 mL of .250 M NaOH. This is my work ?
3. ### chemistry

Which of the following mixtures will be a buffer when dissolved in a liter of water?
4. ### CHEM

Which of the following mixtures will result in the formation of a buffer solution when dissolved in 1.00 L of water?
5. ### General Chemistry

Starting out with 50 mL of 0.20 M NaHCO3, calculate how many mL of 0.50 M NaOH solution to add to make 100 mL of approximately 0.10 M (total) buffer solution with a pH of 10.35. By adding NaOH, some of the NaHCO3 gets converted to …
6. ### chemistry

A chemist wishes to prepare 250mL of a buffer that is pH = 4.50. Beginning with 100mL of 0.12 mol L^(-1) acetic acid and a supply of 0.10 mol L^(-1) NaOH, explain how this could be done. How much 0.20 mol L^(-1) NaOH must be added …
7. ### chemistry

Need help with my pH calculations? 1) 0.10 mol of solid sodium hydrogen carbonate and 0.20 mol of solid sodium carbonate are dissolved in the same beaker of water, transferred to a volumetric ﬂask and made to 250.0 mL. The Ka
8. ### chemistry/science

Not sure if i posted this but Need help with my pH calculations?
9. ### chemistry

200cm^3 of 1.0 mol dm^-3 sulphuric acid,H2SO4 is poured into a 250 cm^3 volumetric flask.Distilled water is then added to make 250 cm^3 of solution. (a)What is the molarity of the diluted acid solution?
10. ### Chemistry

100 g of water is added into 10.6 m NaOH solution. The diluted NaOH solution is then neutralized completely with H2SO4 solution. Density of NaOH is 1.08 g/ml. Calculate: i. the mol fraction of NaOH in the diluted NaOH solution. ii. …

More Similar Questions