3 question was changed to a numbered list, please use \pu for formatting units, see [formatting wiki](https://chemistry.meta.stackexchange.com/a/444) for details edited Apr 28 '18 at 1:43 Gaurang Tandon 5,43288 gold badges3030 silver badges7070 bronze badges It is perhaps difficult to comment on the intended semantics of the option C3, but I think that is the most appropriate answer. AOption 1 and D4 are wrong because of the reasons you cite. To comment upon Boption 2, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $$3.1$$ to a pH of $$4.4$$. For our case, we would end up with a $$\frac{1}{3}\text{M}$$$$\pu{\frac 13 M}$$ solution of HCl$$\ce{HCl}$$, which, assuming complete dissociation and negligible contribution of $$\text{H}^+$$ from water, we end up with a pH of around $$0.48$$ which is well in the no-colour-change region of the pH values. Regarding Coption 3, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' NaCl$$\ce{NaCl}$$ molecules (as in they did not exist bonded to one another before), water molecules are also formed. $$\ce{NaOH + HCl -> NaCl + H2O}$$, i.e. table salt and water It is perhaps difficult to comment on the intended semantics of the option C, but I think that is the most appropriate answer. A and D are wrong because of the reasons you cite. To comment upon B, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $$3.1$$ to a pH of $$4.4$$. For our case, we would end up with a $$\frac{1}{3}\text{M}$$ solution of HCl, which, assuming complete dissociation and negligible contribution of $$\text{H}^+$$ from water, we end up with a pH of around $$0.48$$ which is well in the no-colour-change region of the pH values. Regarding C, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' NaCl molecules (as in they did not exist bonded to one another before), water molecules are also formed. $$\ce{NaOH + HCl -> NaCl + H2O}$$, i.e. table salt and water It is perhaps difficult to comment on the intended semantics of the option 3, but I think that is the most appropriate answer. Option 1 and 4 are wrong because of the reasons you cite. To comment upon option 2, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $$3.1$$ to a pH of $$4.4$$. For our case, we would end up with a $$\pu{\frac 13 M}$$ solution of $$\ce{HCl}$$, which, assuming complete dissociation and negligible contribution of $$\text{H}^+$$ from water, we end up with a pH of around $$0.48$$ which is well in the no-colour-change region of the pH values. Regarding option 3, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' $$\ce{NaCl}$$ molecules (as in they did not exist bonded to one another before), water molecules are also formed. $$\ce{NaOH + HCl -> NaCl + H2O}$$, i.e. table salt and water 2 Showing that water is produced edited Apr 27 '18 at 20:45 DrMoishe Pippik 16.3k1717 silver badges3535 bronze badges It is perhaps difficult to comment on the intended semantics of the option C, but I think that is the most appropriate answer. A and D are wrong because of the reasons you cite. To comment upon B, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $$3.1$$ to a pH of $$4.4$$. For our case, we would end up with a $$\frac{1}{3}\text{M}$$ solution of HCl, which, assuming complete dissociation and negligible contribution of $$\text{H}^+$$ from water, we end up with a pH of around $$0.48$$ which is well in the no-colour-change region of the pH values. Regarding C, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' NaCl molecules (as in they did not exist bonded to one another before), water molecules are also formed. $$\ce{NaOH + HCl -> NaCl + H2O}$$, i.e. table salt and water It is perhaps difficult to comment on the intended semantics of the option C, but I think that is the most appropriate answer. A and D are wrong because of the reasons you cite. To comment upon B, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $$3.1$$ to a pH of $$4.4$$. For our case, we would end up with a $$\frac{1}{3}\text{M}$$ solution of HCl, which, assuming complete dissociation and negligible contribution of $$\text{H}^+$$ from water, we end up with a pH of around $$0.48$$ which is well in the no-colour-change region of the pH values. Regarding C, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' NaCl molecules (as in they did not exist bonded to one another before), water molecules are also formed. It is perhaps difficult to comment on the intended semantics of the option C, but I think that is the most appropriate answer. A and D are wrong because of the reasons you cite. To comment upon B, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $$3.1$$ to a pH of $$4.4$$. For our case, we would end up with a $$\frac{1}{3}\text{M}$$ solution of HCl, which, assuming complete dissociation and negligible contribution of $$\text{H}^+$$ from water, we end up with a pH of around $$0.48$$ which is well in the no-colour-change region of the pH values. Regarding C, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' NaCl molecules (as in they did not exist bonded to one another before), water molecules are also formed. $$\ce{NaOH + HCl -> NaCl + H2O}$$, i.e. table salt and water 1 answered Apr 27 '18 at 20:05 Satwik Pasani 4,17444 gold badges3535 silver badges6565 bronze badges It is perhaps difficult to comment on the intended semantics of the option C, but I think that is the most appropriate answer. A and D are wrong because of the reasons you cite. To comment upon B, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $$3.1$$ to a pH of $$4.4$$. For our case, we would end up with a $$\frac{1}{3}\text{M}$$ solution of HCl, which, assuming complete dissociation and negligible contribution of $$\text{H}^+$$ from water, we end up with a pH of around $$0.48$$ which is well in the no-colour-change region of the pH values. Regarding C, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' NaCl molecules (as in they did not exist bonded to one another before), water molecules are also formed.